Chapter 9, “Quadratics” from Beginning and Intermediate Algebra by Tyler Wallace is available under a Creative Commons Attribution 3.0 Unported license. © 2010.
9.1
Quadratics – Solving with Radicals
Objective: Solve equations with radicals and check for extraneous solu- tions.
Here we look at equations that have roots in the problem. As you might expect, to clear a root we can raise both sides to an exponent. So to clear a square root we can rise both sides to the second power. To clear a cubed root we can raise both sides to a third power. There is one catch to solving a problem with roots in it, sometimes we end up with solutions that do not actually work in the equation. This will only happen if the index on the root is even, and it will not happen all the time. So for these problems it will be required that we check our answer in the original problem. If a value does not work it is called an extraneous solution and not included in the final solution.
When solving a radical problem with an even index: check answers!
Example 442.
7x + 2 √
= 4 Even index!Wewill have to check answers
( 7x+ 2 √
)2 =42 Square both sides, simplify exponents
7x + 2= 16 Solve
− 2 − 2 Subtract 2 fromboth sides 7x = 14 Divide both sides by 7
7 7
x = 2 Need to check answer in original problem
7(2)+ 2 √
= 4 Multiply
14+ 2 √
= 4 Add
16 √
= 4 Square root
4= 4 True! It works!
x = 2 Our Solution
Example 443.
x− 13 √
=− 4 Odd index,we don′t need to check answer ( x− 13 √
)3 =(− 4)3 Cube both sides, simplify exponents x− 1 =− 64 Solve
326
+ 1 +1 Add 1 to both sides
x =− 63 Our Solution
Example 444.
3x +64 √
=− 3 Even index!Wewill have to check answers ( 3x +64 √
) = (− 3)4 Rise both sides to fourth power 3x +6 = 81 Solve
− 6 − 6 Subtract 6 fromboth sides 3x = 75 Divide both sides by 3
3 3
x = 25 Need to check answer in original problem
3(25) + 64 √
=− 3 Multiply 75+64
√ =− 3 Add
814 √
=− 3 Take root 3 =− 3 False, extraneous solution
No Solution Our Solution
If the radical is not alone on one side of the equation we will have to solve for the radical before we raise it to an exponent
Example 445.
x + 4x +1 √
= 5 Even index!Wewill have to check solutions
− x −x Isolate radical by subtracting x fromboth sides 4x +1
√ = 5−x Square both sides
( 4x+ 1 √
)2 =(5− x)2 Evaluate exponents, recal (a− b)2 = a2− 2ab + b2 4x +1 = 25− 10x + x2 Re− order terms 4x +1 =x2− 10x + 25 Make equation equal zero
− 4x− 1 − 4x − 1 Subtract 4x and 1 fromboth sides 0 =x2− 14x + 24 Factor
0= (x− 12)(x− 2) Set each factor equal to zero x− 12=0 or x− 2= 0 Solve each equation + 12+ 12 +2+ 2
x = 12 or x = 2 Need to check answers in original problem
(12) + 4(12) + 1 √
= 5 Checkx =5first
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12+ 48+1 √
= 5 Add
12+ 49 √
= 5 Take root
12+7= 5 Add
19= 5 False, extraneous root
(2) + 4(2) +1 √
= 5 Checkx =2
2+ 8 +1 √
= 5 Add
2+ 9 √
= 5 Take root
2 +3= 5 Add
5= 5 True! Itworks
x = 2 Our Solution
The above example illustrates that as we solve we could end up with an x2 term or a quadratic. In this case we remember to set the equation to zero and solve by factoring. We will have to check both solutions if the index in the problem was even. Sometimes both values work, sometimes only one, and sometimes neither works.
World View Note: The babylonians were the first known culture to solve quadratics in radicals – as early as 2000 BC!
If there is more than one square root in a problem we will clear the roots one at a time. This means we must first isolate one of them before we square both sides.
Example 446.
3x− 8 √
− x√ =0 Even index!Wewill have to check answers + x √
+ x √
Isolate first root by adding x √
to both sides
3x− 8 √
= x √
Square both sides
( 3x− 8 √
)2 = ( x √
)2 Evaluate exponents
3x− 8= x Solve − 3x − 3x Subtract 3x fromboth sides − 8=− 2x Divide both sides by− 2 − 2 − 2
4= x Need to check answer in original
3(4)− 8 √
− 4 √
=0 Multiply
12− 8 √
− 4 √
=0 Subtract
4 √
− 4 √
=0 Take roots
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2− 2=0 Subtract 0=0 True! It works
x =4 Our Solution
When there is more than one square root in the problem, after isolating one root and squaring both sides we may still have a root remaining in the problem. In this case we will again isolate the term with the second root and square both sides. When isolating, we will isolate the term with the square root. This means the square root can be multiplied by a number after isolating.
Example 447.
2x + 1 √
− x√ = 1 Even index!Wewill have to check answers + x √
+ x √
Isolate first root by adding x √
to both sides
2x +1 √
= x √
+ 1 Square both sides
( 2x +1 √
)2 = ( x √
+1)2 Evaluate exponents, recall (a + b)2 = a2 +2ab+ b2
2x +1= x +2 x √
+ 1 Isolate the termwith the root
−x− 1−x − 1 Subtract x and 1 fromboth sides x =2 x
√ Square both sides
(x)2 =(2 x √
)2 Evaluate exponents
x2 =4x Make equation equal zero
− 4x− 4x Subtract x fromboth sides x2− 4x= 0 Factor
x(x− 4)= 0 Set each factor equal to zero x =0 or x− 4= 0 Solve
+4+ 4 Add 4 to both sides of second equation
x = 0 or x= 4 Need to check answers in original
2(0)+ 1 √
− (0) √
= 1 Checkx =0first
1 √
− 0 √
= 1 Take roots
1− 0= 1 Subtract 1= 1 True! Itworks
2(4)+ 1 √
− (4) √
= 1 Checkx =4
8+ 1 √
− 4 √
= 1 Add
9 √
− 4 √
= 1 Take roots
3− 2= 1 Subtract 1= 1 True! Itworks
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x =0 or 4 Our Solution
Example 448.
3x + 9 √
− x + 4 √
=− 1 Even index!Wewill have to check answers + x +4 √
+ x + 4 √
Isolate the first root by adding x + 4 √
3x + 9 √
= x +4 √
− 1 Square both sides ( 3x + 9 √
)2 =( x + 4 √
− 1)2 Evaluate exponents 3x + 9=x + 4− 2 x +4
√ + 1 Combine like terms
3x +9 =x + 5− 2 x + 4 √
Isolate the termwith radical
−x− 5−x− 5 Subtractx and 5 fromboth sides 2x +4=− 2 x + 4
√ Square both sides
(2x + 4)2 =(− 2 x +4 √
)2 Evaluate exponents
4×2 + 16x + 16=4(x + 4) Distribute
4×2 + 16x + 16=4x + 16 Make equation equal zero
− 4x− 16− 4x− 16 Subtract 4x and 16 fromboth sides 4×2 + 12x = 0 Factor
4x(x + 3)= 0 Set each factor equal to zero
4x = 0 or x +3= 0 Solve
4 4 − 3− 3 x =0 or x =− 3 Check solutions in original
3(0)+ 9 √
− (0)+ 4 √
=− 1 Checkx= 0first 9
√ − 4 √
=− 1 Take roots 3− 2=− 1 Subtract
1=− 1 False, extraneous solution
3(− 3) +9 √
− (− 3)+ 4 √
=− 1 Checkx=− 3 − 9+9
√ − (− 3)+ 4 √
=− 1 Add 0
√ − 1 √
=− 1 Take roots 0− 1=− 1 Subtract − 1=− 1 True! Itworks
x =− 3 Our Solution
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9.1 Practice – Solving with Radicals
Solve.
1) 2x + 3 √
− 3= 0
3) 6x− 5 √
− x =0
5) 3+ x= 6x + 13 √
7) 3− 3x √
− 1 =2x
9) 4x + 5 √
− x + 4 √
=2
11) 2x +4 √
− x + 3 √
=1
13) 2x +6 √
− x + 4 √
=1
15) 6− 2x √
− 2x +3 √
= 3
2) 5x +1 √
− 4= 0
4) x + 2 √
− x√ = 2
6) x− 1= 7−x √
8) 2x +2 √
=3 + 2x− 1 √
10) 3x +4 √
− x + 2 √
=2
12) 7x +2 √
− 3x + 6 √
=6
14) 4x− 3 √
− 3x +1 √
= 1
16) 2− 3x √
− 3x +7 √
= 3
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9.2
Quadratics – Solving with Exponents
Objective: Solve equations with exponents using the odd root property and the even root property.
Another type of equation we can solve is one with exponents. As you might expect we can clear exponents by using roots. This is done with very few unex- pected results when the exponent is odd. We solve these problems very straight forward using the odd root property
OddRootProperty: if an = b, then a = b n √
whenn is odd
Example 449.
x5 = 32 Use odd root property
x5 5 √
= 325 √
Simplify roots
x =2 Our Solution
However, when the exponent is even we will have two results from taking an even root of both sides. One will be positive and one will be negative. This is because both 32 = 9 and (− 3)2 = 9. so when solving x2 = 9 we will have two solutions, one positive and one negative: x = 3 and− 3
EvenRoot Property: if an = b, thena = ± bn √
whenn is even
Example 450.
x4 = 16 Use even root property (± )
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x4 4 √
=± 164 √
Simplify roots
x =± 2 Our Solution
World View Note: In 1545, French Mathematicain Gerolamo Cardano pub- lished his book The Great Art, or the Rules of Algebra which included the solu- tion of an equation with a fourth power, but it was considered absurd by many to take a quantity to the fourth power because there are only three dimensions!
Example 451.
(2x + 4)2 = 36 Use even root property (± ) (2x +4)2
√
=± 36 √
Simplify roots
2x + 4=± 6 To avoid sign errors we need two equations 2x + 4= 6 or 2x + 4=− 6 One equation for+ , one equation for− − 4− 4 − 4 − 4 Subtract 4 fromboth sides
2x =2 or 2x =− 10 Divide both sides by 2 2 2 2 2
x =1 or x =− 5 Our Solutions
In the previous example we needed two equations to simplify because when we took the root, our solutions were two rational numbers, 6 and− 6. If the roots did not simplify to rational numbers we can keep the ± in the equation.
Example 452.
(6x− 9)2 = 45 Use even root property (± ) (6x− 9)2
√
=± 45 √
Simplify roots
6x− 9=± 3 5 √
Use one equation because root did not simplify to rational
+ 9 + 9 Add 9 to both sides
6x = 9± 3 5 √
Divide both sides by 6
6 6
x = 9± 3 5
√
6 Simplify, divide each termby 3
x = 3± 5
√
2 Our Solution
333
When solving with exponents, it is important to first isolate the part with the exponent before taking any roots.
Example 453.
(x +4)3− 6= 119 Isolate part with exponent + 6 +6
(x +4)3 = 125 Use odd root property
(x +4)33 √
= 125 √
Simplify roots
x + 4=5 Solve
− 4− 4 Subtract 4 fromboth sides x =1 Our Solution
Example 454.
(6x +1)2 +6 = 10 Isolate partwith exponent
− 6− 6 Subtract 6 fromboth sides (6x + 1)2 =4 Use even root property (± )
(6x +1)2 √
=± 4 √
Simplify roots
6x +1 =± 2 To avoid sign errors,weneed two equations 6x +1 =2 or 6x +1 =− 2 Solve each equation
− 1− 1 − 1 − 1 Subtract 1 fromboth sides 6x = 1 or 6x =− 3 Divide both sides by 6 6 6 6 6
x = 1
6 or x =− 1
2 Our Solution
When our exponents are a fraction we will need to first convert the fractional
exponent into a radical expression to solve. Recall that a m
n = ( an √
) m
. Once we
have done this we can clear the exponent using either the even ( ± ) or odd root property. Then we can clear the radical by raising both sides to an exponent (remember to check answers if the index is even).
Example 455.
(4x +1) 2
5 = 9 Rewrite as a radical expression
( 4x + 15 √
)2 = 9 Clear exponent first with even root property (± ) ( 4x +15 √
)2 √
=± 9 √
Simplify roots
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4x +15 √
=± 3 Clear radical by raising both sides to 5th power ( 4x + 15 √
)5 = (± 3)5 Simplify exponents 4x +1=± 243 Solve, need 2 equations!
4x +1= 243 or 4x +1=− 243 − 1 − 1 − 1 − 1 Subtract 1 fromboth sides
4x = 242 or 4x=− 244 Divide both sides by 4 4 4 4 4
x = 121
2 ,− 61 Our Solution
Example 456.
(3x− 2) 3
4 = 64 Rewrite as radical expression
( 3x− 24 √
)3 = 64 Clear exponent firstwith odd root property
( 3x− 24 √
)33 √
= 643 √
Simplify roots
3x− 24 √
= 4 Even Index!Check answers.
( 3x− 24 √
)4 =44 Raise both sides to 4th power
3x− 2= 256 Solve + 2 + 2 Add 2 to both sides
3x = 258 Divide both sides by 3
3 3
x = 86 Need to check answer in radical form of problem
( 3(86)− 24 √
)3 = 64 Multiply
( 258− 24 √
)3 = 64 Subtract
( 2564 √
)3 = 64 Evaluate root
43 = 64 Evaluate exponent
64= 64 True! It works
x = 86 Our Solution
With rational exponents it is very helpful to convert to radical form to be able to see if we need a ± because we used the even root property, or to see if we need to check our answer because there was an even root in the problem. When checking we will usually want to check in the radical form as it will be easier to evaluate.
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9.2 Practice – Solving with Exponents
Solve.
1) x2 = 75
3) x2 + 5= 13
5) 3×2 + 1= 73
7) (x +2)5 =− 243
9) (2x +5)3− 6= 21
11) (x− 1) 2
3 = 16
13) (2−x) 3
2 = 27
15) (2x− 3) 2
3 = 4
17) (x + 1
2 ) − 2
3 =4
19) (x− 1)− 5
2 = 32
21) (3x− 2) 4
5 = 16
23) (4x +2) 3
5 =− 8
2) x3 =− 8
4) 4×3− 2= 106
6) (x− 4)2 = 49
8) (5x +1)4 = 16
10) (2x +1)2 +3 = 21
12) (x− 1) 3
2 = 8
14) (2x +3) 4
3 = 16
16) (x +3) − 1
3 = 4
18) (x− 1)− 5
3 = 32
20) (x +3) 3
2 =− 8
22) (2x +3) 3
2 = 27
24) (3− 2x) 4
3 =− 81
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9.3
Quadratics – Complete the Square
Objective: Solve quadratic equations by completing the square.
When solving quadratic equations in the past we have used factoring to solve for our variable. This is exactly what is done in the next example.
Example 457.
x2 + 5x +6= 0 Factor
(x + 3)(x +2)= 0 Set each factor equal to zero
x + 3=0 or x +2= 0 Solve each equation
− 3− 3 − 2− 2 x =− 3 or x =− 2 Our Solutions
However, the problem with factoring is all equations cannot be factored. Consider the following equation: x2 − 2x − 7 = 0. The equation cannot be factored, however there are two solutions to this equation, 1 + 2 2
√ and 1 − 2 2
√ . To find these two
solutions we will use a method known as completing the square. When completing the square we will change the quadratic into a perfect square which can easily be solved with the square root property. The next example reviews the square root property.
Example 458.
(x +5)2 = 18 Square root of both sides
(x + 5)2 √
=± 18 √
Simplify each radical
x +5 =± 3 2 √
Subtract 5 fromboth sides
− 5 − 5 x =− 5± 3 2
√ Our Solution
337
To complete the square, or make our problem into the form of the previous example, we will be searching for the third term in a trinomial. If a quadratic is of the form x2 + bx + c, and a perfect square, the third term, c, can be easily
found by the formula (
1
2 · b )2
. This is shown in the following examples, where we
find the number that completes the square and then factor the perfect square.
Example 459.
x2 +8x + c c =
(
1
2 · b )2
and our b =8
(
1
2 · 8 )2
= 42 = 16 The third term to complete the square is 16
x2 + 8x + 16 Our equation as aperfect square, factor
(x +4)2 Our Solution
Example 460.
x2− 7x + c c = (
1
2 · b )2
and our b =7
(
1
2 · 7 )2
=
(
7
2
)2
= 49
4 The third term to complete the square is
49
4
x2− 11x + 49 4
Our equation as aperfect square, factor
(
x− 7 2
)2
Our Solution
Example 461.
x2 + 5
3 x + c c =
(
1
2 · b )2
and our b =8
(
1
2 · 5 3
)2
=
(
5
6
)2
= 25
36 The third term to complete the square is
25
36
338
x2 + 5
3 x+
25
36 Our equation as aperfect square, factor
(
x + 5
6
)2
Our Solution
The process in the previous examples, combined with the even root property, is used to solve quadratic equations by completing the square. The following five steps describe the process used to complete the square, along with an example to demonstrate each step.
Problem 3×2 + 18x− 6=0
1. Separate constant term from variables + 6+6
3×2 + 18x = 6
2. Divide each term by a 3
3 x2 +
18
3 x =
6
3
x2 +6x =2
3. Find value to complete the square: (
1
2 · b )2 ( 1
2 · 6 )2
= 32 =9
4. Add to both sides of equation x2 +6x =2
+9 +9 x2 +6x + 9= 11
5. Factor (x +3)2 = 11
Solve by even root property
(x +3)2 √
=± 11 √
x+ 3=± 11 √
− 3 − 3 x=− 3± 11
√
World View Note: The Chinese in 200 BC were the first known culture group to use a method similar to completing the square, but their method was only used to calculate positive roots.
The advantage of this method is it can be used to solve any quadratic equation. The following examples show how completing the square can give us rational solu- tions, irrational solutions, and even complex solutions.
Example 462.
2×2 + 20x+ 48= 0 Separate constant term fromvaraibles
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− 48− 48 Subtract 24 2×2 + 20x =− 48 Divide by a or 2
2 2 2
x2 + 10x =− 24 Find number to complete the square: (
1
2 · b )2
(
1
2 · 10 )2
= 52 = 25 Add 25 to both sides of the equation
x2 + 10x =− 24 + 25 + 25
x2 + 10x+ 25= 1 Factor
(x +5)2 = 1 Solvewith even root property
(x +5)2 √
=± 1 √
Simplify roots
x + 5=± 1 Subtract 5 fromboth sides − 5− 5
x =− 5± 1 Evaluate x =− 4 or − 6 Our Solution
Example 463.
x2− 3x− 2=0 Separate constant fromvariables + 2+2 Add 2 to both sides
x2− 3x =2 No a,find number to complete the square (
1
2 · b )2
(
1
2 · 3 )2
=
(
3
2
)2
= 9
4 Add
9
4 to both sides,
2
1
(
4
4
)
+ 9
4 =
8
4 +
9
4 =
17
4 Need commondenominator (4) on right
x2− 3x + 9 4
= 8
4 +
9
4 =
17
4 Factor
(
x− 3 2
)2
= 17
4 Solve using the even root property
(
x− 3 2
)2 √
=± 17 4
√
Simplify roots
x− 3 2
= ± 17 √
2 Add
3
2 to both sides,
340
+ 3
2 +
3
2 we already have a common denominator
x = 3± 17
√
2 Our Solution
Example 464.
3×2 =2x− 7 Separate the constant from the variables − 2x− 2x Subtract 2x fromboth sides 3×2− 2x =− 7 Divide each termby a or 3 3 3 3
x2− 2 3 x =− 7
3 Find the number to complete the square
(
1
2 · b )2
(
1
2 · 2 3
)2
=
(
1
3
)2
= 1
9 Add to both sides,
− 7 3
(
3
3
)
+ 1
9 =
− 21 3
+ 1
9 =
− 20 9
get commondenominator on right
x2− 2 3 x+
1
3 =− 20
9 Factor
(
x− 1 3
)2
=− 20 9
Solve using the even root property
(
x− 1 3
)2 √
=± − 20 9
√
Simplify roots
x− 1 3
= ± 2i 5
√
3 Add
1
3 to both sides,
+ 1
3 +
1
3 Already have commondenominator
x = 1± 2i 5
√
3 Our Solution
As several of the examples have shown, when solving by completing the square we will often need to use fractions and be comfortable finding common denominators and adding fractions together. Once we get comfortable solving by completing the square and using the five steps, any quadratic equation can be easily solved.
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9.3 Practice – Complete the Square
Find the value that completes the square and then rewrite as a perfect square.
1) x2− 30x +__ 3) m2− 36m +__ 5) x2− 15x + __ 7) y2− y +__
2) a2− 24a +__ 4) x2− 34x +__ 6) r2− 1
9 r +__
8) p2− 17p +__ Solve each equation by completing the square.
9) x2− 16x + 55=0 11) v2− 8v + 45= 0 13) 6×2 + 12x + 63= 0
15) 5k2− 10k + 48=0 17) x2 + 10x− 57=4 19) n2− 16n + 67= 4 21) 2×2 + 4x + 38=− 6 23) 8b2 + 16b− 37=5 25) x2 =− 10x− 29 27) n2 =− 21+ 10n 29) 3k2 +9= 6k
31) 2×2 + 63=8x
33) p2− 8p =− 55 35) 7n2−n +7= 7n +6n2
37) 13b2 + 15b + 44=− 5+7b2 +3b 39) 5×2 + 5x =− 31− 5x 41) v2 + 5v + 28=0
43) 7×2− 6x + 40=0 45) k2− 7k + 50= 3 47) 5×2 + 8x− 40=8 49) m2 =− 15+9m 51) 8r2 + 10r =− 55 53) 5n2− 8n + 60=− 3n +6+ 4n2
55) − 2×2 + 3x− 5=− 4×2
10) n2− 8n− 12= 0 12) b2 + 2b+ 43= 0
14) 3×2− 6x + 47=0 16) 8a2 + 16a− 1= 0 18) p2− 16p− 52= 0 20) m2− 8m− 3 =6 22) 6r2 + 12r − 24=− 6 24) 6n2− 12n− 14=4 26) v2 = 14v + 36
28) a2− 56=− 10a 30) 5×2 =− 26+ 10x 32) 5n2 =− 10n + 15 34) x2 + 8x+ 15= 8
36) n2 +4n = 12
38) − 3r2 + 12r + 49=− 6r2
40) 8n2 + 16n = 64
42) b2 + 7b− 33=0 44) 4×2 + 4x + 25= 0
46) a2− 5a + 25= 3 48) 2p2− p + 56=− 8 50) n2−n =− 41 52) 3×2− 11x =− 18 54) 4b2− 15b + 56=3b2
56) 10v2− 15v = 27+ 4v2− 6v
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9.4
Quadratics – Quadratic Formula
Objective: Solve quadratic equations by using the quadratic formula.
The general from of a quadratic is ax2 + bx + c = 0. We will now solve this for- mula for x by completing the square
Example 465.
ax2 + bc + c =0 Separate constant fromvariables
− c− c Subtract c fromboth sides ax2 + bx =− c Divide each termby a a a a
x2 + b
a x =
− c a
Find the number that completes the square (
1
2 · b a
)2
=
(
b
2a
)2
= b2
4a2 Add to both sides,
b2
4a2 − c
a
(
4a
4a
)
= b2
4a2 − 4ac
4a2 =
b2− 4ac 4a2
Get common denominator on right
x2 + b
a x +
b2
4a2 =
b2
4a2 − 4ac
4a2 =
b2− 4ac 4a2
Factor
(
x + b
2a
)2
= b2− 4ac
4a2 Solve using the even root property
(
x + b
2a
)2 √
=± b 2− 4ac 4a2
√
Simplify roots
x + b
2a =
± b2− 4ac √
2a Subtract
b
2a fromboth sides
x = − b± b2− 4ac
√
2a Our Solution
This solution is a very important one to us. As we solved a general equation by completing the square, we can use this formula to solve any quadratic equation. Once we identify what a, b, and c are in the quadratic, we can substitute those
343
values into x = − b± b2− 4ac
√
2a and we will get our two solutions. This formula is
known as the quadratic fromula
Quadratic Formula: if ax2 + bx + c = 0 then x = − b ± b2 − 4ac
√
2a
World View Note: Indian mathematician Brahmagupta gave the first explicit formula for solving quadratics in 628. However, at that time mathematics was not done with variables and symbols, so the formula he gave was, “To the absolute number multiplied by four times the square, add the square of the middle term; the square root of the same, less the middle term, being divided by twice the square is the value.” This would translate to
4ac+ b2 √
− b 2a
as the solution to the equation ax2 + bx = c.
We can use the quadratic formula to solve any quadratic, this is shown in the fol- lowing examples.
Example 466.
x2 +3x + 2=0 a= 1, b = 3, c =2, use quadratic formula
x = − 3± 32− 4(1)(2)
√
2(1) Evaluate exponent andmultiplication
x = − 3± 9− 8
√
2 Evaluate subtraction under root
x = − 3± 1
√
2 Evaluate root
x = − 3± 1
2 Evaluate± to get two answers
x= − 2 2
or − 4 2
Simplify fractions
x =− 1 or − 2 Our Solution
As we are solving using the quadratic formula, it is important to remember the equation must fist be equal to zero.
Example 467.
25×2 = 30x + 11 First set equal to zero
− 30x− 11 − 30x− 11 Subtract 30x and 11 fromboth sides 25×2− 30x− 11=0 a= 25, b =− 30, c=− 11,use quadratic formula
x = 30± (− 30)2− 4(25)(− 11)
√
2(25) Evaluate exponent andmultiplication
344
x = 30± 900+ 1100
√
50 Evaluate addition inside root
x = 30± 2000
√
50 Simplify root
x = 30± 20 5
√
50 Reduce fraction by dividing each termby 10
x = 3± 2 5
√
5 Our Solution
Example 468.
3×2 + 4x+ 8= 2×2 +6x− 5 First set equation equal to zero − 2×2− 6x+ 5 − 2×2− 6x + 5 Subtract 2×2 and 6x and add 5
x2− 2x + 13=0 a =1, b =− 2, c= 13, use quadratic formula
x = 2± (− 2)2− 4(1)(13)
√
2(1) Evaluate exponent andmultiplication
x = 2± 4− 52
√
2 Evaluate subtraction inside root
x = 2± − 48
√
2 Simplify root
x = 2± 4i 3
√
2 Reduce fraction by dividing each termby 2
x = 1± 2i 3 √
Our Solution
When we use the quadratic formula we don’t necessarily get two unique answers. We can end up with only one solution if the square root simplifies to zero.
Example 469.
4×2− 12x +9= 0 a = 4, b =− 12, c= 9,use quadratic formula
x = 12± (− 12)2− 4(4)(9)
√
2(4) Evaluate exponents andmultiplication
x = 12± 144− 144
√
8 Evaluate subtraction inside root
x = 12± 0
√
8 Evaluate root
x = 12± 0
8 Evaluate±
x = 12
8 Reduce fraction
x= 3
2 Our Solution
345
If a term is missing from the quadratic, we can still solve with the quadratic for- mula, we simply use zero for that term. The order is important, so if the term with x is missing, we have b =0, if the constant term is missing, we have c= 0.
Example 470.
3×2 +7 =0 a = 3, b = 0(missing term), c =7
x = − 0± 02− 4(3)(7)
√
2(3) Evaluate exponnets andmultiplication, zeros not needed
x = ± − 84 √
6 Simplify root
x = ± 2i 21
√
6 Reduce,dividing by 2
x = ± i 21
√
3 Our Solution
We have covered three different methods to use to solve a quadratic: factoring, complete the square, and the quadratic formula. It is important to be familiar with all three as each has its advantage to solving quadratics. The following table walks through a suggested process to decide which method would be best to use for solving a problem.
1. If it can easily factor, solve by factoring x2− 5x + 6=0 (x− 2)(x− 3) =0 x =2 or x = 3
2. If a= 1 and b is even, complete the square
x2 + 2x =4 (
1
2 · 2 )2
= 12 =1
x2 + 2x +1= 5 (x +1)2 = 5
x +1 =± 5 √
x =− 1± 5 √
3. Otherwise, solve by the quadratic formula
x2− 3x + 4=0 x =
3± (− 3)2− 4(1)(4) √
2(1)
x = 3± i 7
√
2
The above table is mearly a suggestion for deciding how to solve a quadtratic. Remember completing the square and quadratic formula will always work to solve any quadratic. Factoring only woks if the equation can be factored.
346
9.4 Practice – Quadratic Formula
Solve each equation with the quadratic formula.
1) 4a2 +6 =0
3) 2×2− 8x− 2 =0
5) 2m2− 3= 0
7) 3r2− 2r − 1=0
9) 4n2− 36= 0
11) v2− 4v − 5=− 8
13) 2a2 +3a + 14=6
15) 3k2 +3k − 4= 7
17) 7×2 + 3x− 16=− 2
19) 2p2 + 6p− 16=4
21) 3n2 +3n =− 3
23) 2×2 =− 7x + 49
25) 5×2 = 7x + 7
27) 8n2 =− 3n− 8
29) 2×2 + 5x =− 3
31) 4a2− 64= 0
33) 4p2 + 5p− 36=3p2
35) − 5n2− 3n− 52= 2− 7n2
37) 7r2− 12=− 3r
39) 2n2− 9= 4
2) 3k2 +2= 0
4) 6n2− 1= 0
6) 5p2 + 2p+ 6=0
8) 2×2− 2x− 15= 0
10) 3b2 + 6=0
12) 2×2 + 4x + 12= 8
14) 6n2− 3n +3=− 4
16) 4×2− 14=− 2
18) 4n2 +5n = 7
20) m2 + 4m− 48=− 3
22) 3b2− 3= 8b
24) 3r2 + 4=− 6r
26) 6a2 =− 5a + 13
28) 6v2 = 4+ 6v
30) x2 = 8
32) 2k2 +6k − 16= 2k
34) 12×2 +x + 7= 5×2 +5x
36) 7m2− 6m + 6=−m
38) 3×2− 3= x2
40) 6b2 = b2 +7− b
347
9.5
Quadratics – Build Quadratics From Roots
Objective: Find a quadratic equation that has given roots using reverse factoring and reverse completing the square.
Up to this point we have found the solutions to quadratics by a method such as factoring or completing the square. Here we will take our solutions and work backwards to find what quadratic goes with the solutions.
We will start with rational solutions. If we have rational solutions we can use fac- toring in reverse, we will set each solution equal to x and then make the equation equal to zero by adding or subtracting. Once we have done this our expressions will become the factors of the quadratic.
Example 471.
The solutions are 4 and− 2 Set each solution equal to x x =4 or x =− 2 Make each equation equal zero
− 4− 4 +2 + 2 Subtract 4 fromfirst, add 2 to second x− 4= 0 or x + 2= 0 These expressions are the factors
(x− 4)(x + 2)= 0 FOIL x2 +2x− 4x− 8 Combine like terms x2− 2x− 8= 0 Our Solution
If one or both of the solutions are fractions we will clear the fractions by multi- plying by the denominators.
Example 472.
The solution are 2
3 and
3
4 Set each solution equal tox
x = 2
3 or x =
3
4 Clear fractions bymultiplying by denominators
3x =2 or 4x =3 Make each equation equal zero
− 2− 2 − 3− 3 Subtract 2 from the first, subtract 3 from the second 3x− 2= 0 or 4x− 3=0 These expressions are the factors
(3x− 2)(4x− 3) =0 FOIL 12×2− 9x− 8x + 6=0 Combine like terms
348
12×2− 17x + 6=0 Our Solution
If the solutions have radicals (or complex numbers) then we cannot use reverse factoring. In these cases we will use reverse completing the square. When there are radicals the solutions will always come in pairs, one with a plus, one with a minus, that can be combined into “one” solution using ± . We will then set this solution equal to x and square both sides. This will clear the radical from our problem.
Example 473.
The solutions are 3 √
and− 3 √
Write as ′′one′′ expression equal tox
x =± 3 √
Square both sides
x2 =3 Make equal to zero
− 3− 3 Subtract 3 fromboth sides x2− 3=0 Our Solution
We may have to isolate the term with the square root (with plus or minus) by adding or subtracting. With these problems, remember to square a binomial we use the formula (a + b)2 = a2 + 2ab + b2
Example 474.
The solutions are 2− 5 2 √
and 2+ 5 2 √
Write as ′′one′′ expression equal tox
x =2± 5 2 √
Isolate the square root term
− 2− 2 Subtract 2 fromboth sides x− 2=± 5 2
√ Square both sides
x2− 4x+ 4= 25 · 2 x2− 4x +4 = 50 Make equal to zero
− 50− 50 Subtract 50 x2− 4x− 46= 0 Our Solution
World View Note: Before the quadratic formula, before completing the square, before factoring, quadratics were solved geometrically by the Greeks as early as 300 BC! In 1079 Omar Khayyam, a Persian mathematician solved cubic equations geometrically!
If the solution is a fraction we will clear it just as before by multiplying by the denominator.
349
Example 475.
The solutions are 2 + 3
√
4 and
2− 3 √
4 Write as ′′one′′ expresion equal tox
x = 2± 3
√
4 Clear fraction bymultiplying by 4
4x = 2± 3 √
Isolate the square root term
− 2− 2 Subtract 2 fromboth sides 4x− 2=± 3
√ Square both sides
16×2− 16x + 4=3 Make equal to zero − 3− 3 Subtract 3
16×2− 16x + 1=0 Our Solution
The process used for complex solutions is identical to the process used for radi- cals.
Example 476.
The solutions are 4− 5i and 4+ 5i Write as ′′one′′ expression equal tox x =4± 5i Isolate the i term
− 4− 4 Subtract 4 fromboth sides x− 4=± 5i Square both sides
x2− 8x + 16= 25i2 i2 =− 1 x2− 8x + 16=− 25 Make equal to zero
+ 25 + 25 Add 25 to both sides
x2− 8x + 41=0 Our Solution
Example 477.
The solutions are 3− 5i
2 and
3+ 5i
2 Write as ′′one′′ expression equal to x
x = 3± 5i
2 Clear fraction bymultiplying by denominator
2x =3± 5i Isolate the i term − 3− 3 Subtract 3 fromboth sides
2x− 3=± 5i Square both sides 4×2− 12x + 9=5i2 i2 =− 1
4×2− 12x +9 =− 25 Make equal to zero + 25 + 25 Add 25 to both sides
4×2− 12x + 34=0 Our Solution
350
9.5 Practice – Build Quadratics from Roots
From each problem, find a quadratic equation with those numbers as its solutions.
1) 2, 5
3) 20, 2
5) 4, 4
7) 0, 0
9) − 4, 11
11) 3
4 ,
1
4
13) 1
2 ,
1
3
15) 3
7 , 4
17) − 1 3 ,
5
6
19) − 6, 1 9
21) ± 5
23) ± 1 5
25) ± 11 √
27) ± 3 √
4
29) ± i 13 √
31) 2± 6 √
33) 1± 3i
35) 6± i 3 √
37) − 1± 6
√
2
39) 6± i 2
√
8
2) 3, 6
4) 13, 1
6) 0, 9
8) − 2,− 5
10) 3, − 1
12) 5
8 ,
5
7
14) 1
2 ,
2
3
16) 2, 2
9
18) 5
3 ,− 1
2
20) − 2 5 , 0
22) ± 1
24) ± 7 √
26) ± 2 3 √
28) ± 11i
30) ± 5i 2 √
32) − 3± 2 √
34) − 2± 4i
36) − 9± i 5 √
38) 2± 5i
3
40) − 2± i 15
√
2
351
9.6
Quadratics – Quadratic in Form
Objective: Solve equations that are quadratic in form by substitution to create a quadratic equation.
We have seen three different ways to solve quadratics: factoring, completing the square, and the quadratic formula. A quadratic is any equation of the form 0 = ax2 + bx + c, however, we can use the skills learned to solve quadratics to solve problems with higher (or sometimes lower) powers if the equation is in what is called quadratic form.
Quadratic Form: 0 = axm+ bxn + cwherem = 2n
An equation is in quadratic form if one of the exponents on a variable is double the exponent on the same variable somewhere else in the equation. If this is the case we can create a new variable, set it equal to the variable with smallest expo- nent. When we substitute this into the equation we will have a quadratic equa- tion we can solve.
World View Note: Arab mathematicians around the year 1000 were the first to use this method!
Example 478.
x4− 13×2 + 36=0 Quadratic form, one exponent, 4, double the other, 2 y =x2 Newvariableequaltothevariablewithsmallerexponent
y2 =x4 Square both sides
y2− 13y + 36=0 Substitute y forx2 and y2 for x4 (y − 9)(y − 4)=0 Solve.Wecan solve this equation by factoring
y − 9 =0 or y − 4=0 Set each factor equal to zero +9 +9 + 4+4 Solve each equation
y = 9 or y =4 Solutions for y,needx.Wewill use y =x2 equation
9= x2 or 4=x2 Substitute values for y
± 9 √
= x2 √
or ± 4 √
= x2 √
Solve using the even root property, simplify roots
x =± 3,± 2 Our Solutions
When we have higher powers of our variable, we could end up with many more solutions. The previous equation had four unique solutions.
352
Example 479.
a−2− a−1− 6=0 Quadratic form, one exponent,− 2, is double the other,− 1 b = a−1 Makeanewvariableequaltothevariablewith lowestexponent
b2 = a−2 Square both sides
b2− b− 6=0 Substitute b2 for a−2 and b for a−1 (b− 3)(b +2) =0 Solve.Wewill solve by factoring
b− 3=0 or b + 2=0 Set each factor equal to zero +3 +3 − 2− 2 Solve each equation
b= 3 or b =− 2 Solutions for b, still need a, substitute into b = a−1 3 = a−1 or − 2= a−1 Raise both sides to− 1power
3−1 = a or (− 2)−1 = a Simplify negative exponents a=
1
3 ,− 1
2 Our Solution
Just as with regular quadratics, these problems will not always have rational solu- tions. We also can have irrational or complex solutions to our equations.
Example 480.
2×4 +x2 =6 Make equation equal to zero
− 6− 6 Subtract 6 fromboth sides 2×4 +x2− 6 =0 Quadratic form, one exponent, 4,double the other, 2
y =x2 Newvariableequalvariablewithsmallestexponent
y2 =x4 Square both sides
2y2 + y − 6 =0 Solve.Wewill factor this equation (2y − 3)(y +2) =0 Set each factor equal to zero
2y − 3= 0 or y +2 =0 Solve each equation + 3+ 3 − 2− 2
2y =3 or y =− 2 2 2
y = 3
2 or y =− 2 Wehave y, still need x. Substitute into y =x2
3
2 =x2 or − 2 =x2 Square root of each side
± 3 2
√
= x2 √
or ± − 2 √
= x2 √
Simplify each root, rationalize denominator
x = ± 6 √
2 ,± i 2
√ Our Solution
353
When we create a new variable for our substitution, it won’t always be equal to just another variable. We can make our substitution variable equal to an expres- sion as shown in the next example.
Example 481.
3(x− 7)2− 2(x− 7)+ 5=0 Quadratic form y =x− 7 Define newvariable
y2 = (x− 7)2 Square both sides 3y2− 2y + 5=0 Substitute values into original
(3y − 5)(y +1) =0 Factor 3y − 5= 0 or y + 1=0 Set each factor equal to zero
+5+ 5 − 1− 1 Solve each equation 3y =5 or y =− 1 3 3
y = 5
3 or y =− 1 Wehave y,we still needx.
5
3 =x− 7 or − 1 =x− 7 Substitute into y =x− 7
+ 21
3 +7 +7 +7 Add 7.Use common denominator as needed
x = 26
3 , 6 Our Solution
Example 482.
(x2− 6x)2 =7(x2− 6x)− 12 Make equation equal zero − 7(x2− 6x)+ 12− 7(x2− 6x) + 12 Move all terms to left
(x2− 6x)2− 7(x2− 6x) + 12= 0 Quadratic form y = x2− 6x Make newvariable
y2 =(x2− 6x)2 Square both sides y2− 7y + 12= 0 Substitute into original equation
(y − 3)(y − 4)= 0 Solve by factoring y − 3= 0 or y − 4= 0 Set each factor equal to zero
+3+ 3 +4+ 4 Solve each equation
y =3 or y =4 Wehave y, still needx.
3 =x2− 6x or 4= x3− 6x Solve each equation, complete the square (
1
2 · 6 )2
= 32 = 9 Add 9 to both sides of each equation
12= x2− 6x + 9 or 13=x2− 6x+ 9 Factor
354
12= (x− 3)2 or 13=(x− 3)2 Use even root property ± 12 √
= (x− 3)2 √
or ± 13 √
= (x− 3)2 √
Simplify roots
± 2 3 √
= x− 3 or ± 13 √
= x− 3 Add 3 to both sides +3 +3 + 3 + 3
x = 3± 2 3 √
, 3± 13 √
Our Solution
The higher the exponent, the more solution we could have. This is illustrated in the following example, one with six solutions.
Example 483.
x6− 9×3 + 8= 0 Quadratic form, one exponent, 6,double the other, 3 y =x3 Newvariableequaltovariablewithlowestexponent
y2 =x6 Square both sides
y2− 9y + 8= 0 Substitute y2 forx6 and y for x3 (y − 1)(y − 8)= 0 Solve.Wewill solve by factoring.
y − 1=0 or y − 8= 0 Set each factor equal to zero +1 +1 + 8+ 8 Solve each equation
y = 1 or y = 8 Solutions for y,we need x.Substitute into y = x3
x3 = 1 or x3 = 8 Set each equation equal to zero
− 1− 1 − 8− 8 x3− 1=0 or x3− 8= 0 Factor each equation,difference of cubes (x− 1)(x2 + x +1)= 0 Firstequation factored.Seteachfactorequaltozero
x− 1= 0 or x2 +x + 1= 0 First equation is easy to solve + 1+ 1
x = 1 First solution
− 1± 12− 4(1)(1) √
2(1) =
1± i 3 √
2 Quadratic formula on second factor
(x− 2)(x2 +2x + 4)= 0 Factor the seconddifference of cubes x− 2= 0 or x2 +2x + 4= 0 Set each factor equal to zero.
+2+ 2 First equation is easy to solve
x =2 Our fourth solution
− 2± 22− 4(1)(4) √
2(1) =− 1± i 3
√ Quadratic formula on second factor
x = 1, 2, 1± i 3
√
2 ,− 1± i 3
√ Our final six solutions
355
9.6 Practice – Quadratic in Form
Solve each of the following equations. Some equations will have complex roots.
1) x4− 5×2 +4= 0
3) m4− 7m2− 8=0
5) a4− 50a2 + 49=0
7) x4− 25×2 + 144= 0
9) m4− 20m2 + 64= 0
11) z6− 216= 19z3
13) 6z4− z2 = 12
15) x 2
3 − 35=2x 1
3
17) y−6 + 7y−3 =8
19) x4− 2×2− 3=0
21) 2×4− 5×2 + 2= 0
23) x4− 9×2 +8= 0
25) 8×6− 9×3 + 1= 0
27) x8− 17×4 + 16= 0
29) (y + b)2− 4(y + b)= 21
31) (y + 2)2− 6(y + 2)= 16
33) (x− 3)2− 2(x− 3)= 35
35) (r − 1)2− 8(r − 1) = 20
37) 3(y + 1)2− 14(y + 1) =5
39) (3×2− 2x)2 +5= 6(3×2− 2x)
41) 2(3x +1) 2
3 − 5(3x + 1) 1
3 = 88
43) (x2 + 2x)2− 2(x2 +2x)= 3
45) (2×2−x)2− 4(2×2−x)+ 3=0
2) y4− 9y2 + 20= 0
4) y4− 29y2 + 100= 0
6) b4− 10b2 + 9= 0
8) y4− 40y2 + 144= 0
10) x6− 35×3 + 216= 0
12) y4− 2y2 = 24
14) x−2−x−1− 12= 0
16) 5y−2− 20= 21y−1
18) x4− 7×2 + 12= 0
20) x4 + 7×2 + 10=0
22) 2×4−x2− 3=0
24) x6− 10×3 + 16= 0
26) 8×6 + 7×3− 1= 0
28) (x− 1)2− 4(x− 1)= 5
30) (x +1)2 +6(x + 1)+ 9=0
32) (m− 1)2− 5(m− 1)= 14
34) (a +1)2 +2(a− 1)= 15
36) 2(x− 1)2− (x− 1)= 3
38) (x2− 3)2− 2(x2− 3)=3
40) (x2 + x +3)2 + 15=8(x2 + x +3)
42) (x2 + x)2− 8(x2 +x)+ 12=0
44) (2×2 + 3x)2 =8(2×2 +3x)+ 9
46) (3×2− 4x)2 =3(3×2− 4x) + 4
356
9.7
Quadratics – Rectangles
Objective: Solve applications of quadratic equations using rectangles.
An application of solving quadratic equations comes from the formula for the area of a rectangle. The area of a rectangle can be calculated by multiplying the width by the length. To solve problems with rectangles we will first draw a picture to represent the problem and use the picture to set up our equation.
Example 484.
The length of a rectangle is 3 more than the width. If the area is 40 square inches, what are the dimensions?
40 x Wedonot know thewidth, x.
x +3 Length is 4more, orx +4, and area is 40.
357
x(x + 3)= 40 Multiply length bywidth to get area
x2 +3x = 40 Distribute
− 40− 40 Make equation equal zero x2 + 3x− 40=0 Factor
(x− 5)(x + 8)=0 Set each factor equal to zero x− 5=0 or x + 8=0 Solve each equation +5+ 5 − 8− 8
x =5 or x =− 8 Our x is awidth, cannot be negative. (5)+ 3=8 Length isx +3, substitute 5 for x to find length
5 in by 8 in Our Solution
The above rectangle problem is very simple as there is only one rectangle involved. When we compare two rectangles, we may have to get a bit more cre- ative.
Example 485.
If each side of a square is increased by 6, the area is multiplied by 16. Find the side of the original square.
x2 x Square has all sides the same length
x Area is found bymultiplying length bywidth
16×2 x +6 Each side is increased by 6,
x +6 Area is 16 times original area
(x + 6)(x +6) = 16×2 Multiply length bywidth to get area
x2 + 12x + 36= 16×2 FOIL
− 16×2 − 16×2 Make equation equal zero − 15×2 + 12x + 36=0 Divide each termby− 1, changes the signs 15×2− 12x− 36=0 Solve using the quadratic formula
x= 12± (− 12)2− 4(15)(− 36)
√
2(15) Evaluate
x = 16± 2304
√
30
x = 16± 48
30 Can ′thave anegative solution,wewill only add
x = 60
30 =2 Ourx is the original square
358
2 Our Solution
Example 486.
The length of a rectangle is 4 ft greater than the width. If each dimension is increased by 3, the new area will be 33 square feet larger. Find the dimensions of the original rectangle.
x(x +4) x Wedon ′t knowwidth, x, length is 4more, x +4
x +4 Area is foundbymultiplying length bywidth
x(x +4)+ 33 x +3 Increase each side by 3. width becomesx + 3, length x + 4+3 =x +7
x + 7 Area is 33more than original, x(x +4) + 33
(x +3)(x + 7)=x(x + 4) + 33 Set up equation, length timeswidth is area
x2 + 10x + 21=x2 + 4x + 33 Subtract x2 fromboth sides
−x2 −x2 10x + 21= 4x + 33 Move variables to one side
− 4x − 4x Subtract 4x from each side 6x + 21= 33 Subtract 21 fromboth sides
− 21− 21 6x = 12 Divide both sides by 6
6 6
x =2 x is thewidth of the original
(2)+ 4=6 x +4 is the length. Substitute 2 to find
2 ft by 6ft Our Solution
From one rectangle we can find two equations. Perimeter is found by adding all the sides of a polygon together. A rectangle has two widths and two lengths, both the same size. So we can use the equation P = 2l + 2w (twice the length plus twice the width).
Example 487.
The area of a rectangle is 168 cm2. The perimeter of the same rectangle is 52 cm. What are the dimensions of the rectangle?
x Wedon ′t know anything about length orwidth
y Use two variables, x and y
xy = 168 Length timeswidth gives the area.
2x +2y = 52 Also use perimeter formula.
− 2x − 2x Solve by substitution, isolate y 2y =− 2x + 52 Divide each termby 2
359
2 2 2
y =−x + 26 Substitute into area equation x(−x + 26)= 168 Distribute −x2 + 26x = 168 Divide each termby− 1, changing all the signs x2− 26x =− 168 Solve by completing the square.
(
1
2 · 26 )2
= 132 = 169 Find number to complete the square:
(
1
2 · b )2
x2− 26x + 324=1 Add 169 to both sides (x− 13)2 =1 Factor x− 13=± 1 Square root both sides
+ 13 + 13
x = 13± 1 Evaluate x = 14 or 12 Two options for first side.
y =− (14)+ 26= 12 Substitute 14 into y =− x + 26 y =− (12)+ 26= 14 Substitute 12 into y =− x + 26
Both are the same rectangle, variables switched!
12 cmby 14cm Our Solution
World View Note: Indian mathematical records from the 9th century demon- strate that their civilization had worked extensivly in geometry creating religious alters of various shapes including rectangles.
Another type of rectangle problem is what we will call a “frame problem”. The idea behind a frame problem is that a rectangle, such as a photograph, is centered inside another rectangle, such as a frame. In these cases it will be important to rememember that the frame extends on all sides of the rectangle. This is shown in the following example.
Example 488.
An 8 in by 12 in picture has a frame of uniform width around it. The area of the frame is equal to the area of the picture. What is the width of the frame?
8
12 12+2x Drawpicture, picture if 8by 10 If frame haswidthx, on both sides,weadd 2x
8+ 2x
8 · 12= 96 Area of the picture, length timeswidth 2 · 96= 192 Frameisthesameasthepicture.Totalarea isdoublethis.
(12+2x)(8+ 2x) = 192 Area of everything, length timeswidth
96+ 24x + 16x +4×2 = 192 FOIL
4×2 + 40x + 96= 192 Combine like terms
− 192− 192 Make equation equal to zero by subtracting 192 4×2 + 40x− 96=0 Factor outGCF of 4
360
4(x2 + 10x− 24) = 0 Factor trinomial 4(x− 2)(x + 12) = 0 Set each factor equal to zero
x− 2= 0 or x + 12=0 Solve each equation +2+ 2 − 12− 12
x= 2 or − 12 Can ′t have negative framewidth. 2 inches Our Solution
Example 489.
A farmer has a field that is 400 rods by 200 rods. He is mowing the field in a spiral pattern, starting from the outside and working in towards the center. After an hour of work, 72% of the field is left uncut. What is the size of the ring cut around the outside?
400− 2x
200− 2x 200 Drawpicture, outside is 200 by 400 If frame haswidthx on both sides, subtract 2x from each side to get center
400
400 · 200= 80000 Area of entire field, length timeswidth 80000 · (0.72) = 57600 Area of center,multiply by 28% as decimal
(400− 2x)(200− 2x) = 57600 Area of center, length timeswidth 80000− 800x− 400x +4×2 = 57600 FOIL
4×2− 1200x + 80000= 57600 Combine like terms − 57600− 57600 Make equation equal zero
4×2− 1200x + 22400=0 Factor outGCFof 4 4(x2− 300x + 5600) = 0 Factor trinomial 4(x− 280)(x− 20) = 0 Set each factor equal to zero
x− 280= 0 or x− 20=0 Solve each equation + 280+ 280 + 20+ 20
x = 280 or 20 The field is only 200 rodswide,
Can ′t cut 280 off two sides!
20 rods Our Solution
For each of the frame problems above we could have also completed the square or use the quadratic formula to solve the trinomials. Remember that completing the square or the quadratic formula always will work when solving, however, factoring only works if we can factor the trinomial.
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9.7 Practice – Rectangles
1) In a landscape plan, a rectangular flowerbed is designed to be 4 meters longer than it is wide. If 60 square meters are needed for the plants in the bed, what should the dimensions of the rectangular bed be?
2) If the side of a square is increased by 5 the area is multiplied by 4. Find the side of the original square.
3) A rectangular lot is 20 yards longer than it is wide and its area is 2400 square yards. Find the dimensions of the lot.
4) The length of a room is 8 ft greater than it is width. If each dimension is increased by 2 ft, the area will be increased by 60 sq. ft. Find the dimensions of the rooms.
5) The length of a rectangular lot is 4 rods greater than its width, and its area is 60 square rods. Find the dimensions of the lot.
6) The length of a rectangle is 15 ft greater than its width. If each dimension is decreased by 2 ft, the area will be decreased by 106 ft2. Find the dimensions.
7) A rectangular piece of paper is twice as long as a square piece and 3 inches wider. The area of the rectangular piece is 108 in2. Find the dimensions of the square piece.
8) A room is one yard longer than it is wide. At 75c per sq. yd. a covering for the floor costs S31.50. Find the dimensions of the floor.
9) The area of a rectangle is 48 ft2 and its perimeter is 32 ft. Find its length and width.
10) The dimensions of a picture inside a frame of uniform width are 12 by 16 inches. If the whole area (picture and frame) is 288 in2, what is the width of the frame?
11) A mirror 14 inches by 15 inches has a frame of uniform width. If the area of the frame equals that of the mirror, what is the width of the frame.
12) A lawn is 60 ft by 80 ft. How wide a strip must be cut around it when mowing the grass to have cut half of it.
13) A grass plot 9 yards long and 6 yards wide has a path of uniform width around it. If the area of the path is equal to the area of the plot, determine the width of the path.
14) A landscape architect is designing a rectangular flowerbed to be border with 28 plants that are placed 1 meter apart. He needs an inner rectangular space in the center for plants that must be 1 meter from the border of the bed and
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that require 24 square meters for planting. What should the overall dimensions of the flowerbed be?
15) A page is to have a margin of 1 inch, and is to contain 35 in2 of painting. How large must the page be if the length is to exceed the width by 2 inches?
16) A picture 10 inches long by 8 inches wide has a frame whose area is one half the area of the picture. What are the outside dimensions of the frame?
17) A rectangular wheat field is 80 rods long by 60 rods wide. A strip of uniform width is cut around the field, so that half the grain is left standing in the form of a rectangular plot. How wide is the strip that is cut?
18) A picture 8 inches by 12 inches is placed in a frame of uniform width. If the area of the frame equals the area of the picture find the width of the frame.
19) A rectangular field 225 ft by 120 ft has a ring of uniform width cut around the outside edge. The ring leaves 65% of the field uncut in the center. What is the width of the ring?
20) One Saturday morning George goes out to cut his lot that is 100 ft by 120 ft. He starts cutting around the outside boundary spiraling around towards the center. By noon he has cut 60% of the lawn. What is the width of the ring that he has cut?
21) A frame is 15 in by 25 in and is of uniform width. The inside of the frame leaves 75% of the total area available for the picture. What is the width of the frame?
22) A farmer has a field 180 ft by 240 ft. He wants to increase the area of the field by 50% by cultivating a band of uniform width around the outside. How wide a band should he cultivate?
23) The farmer in the previous problem has a neighber who has a field 325 ft by 420 ft. His neighbor wants to increase the size of his field by 20% by cultivating a band of uniform width around the outside of his lot. How wide a band should his neighbor cultivate?
24) A third farmer has a field that is 500 ft by 550 ft. He wants to increase his field by 20%. How wide a ring should he cultivate around the outside of his field?
25) Donna has a garden that is 30 ft by 36 ft. She wants to increase the size of the garden by 40%. How wide a ring around the outside should she cultivate?
26) A picture is 12 in by 25 in and is surrounded by a frame of uniform width. The area of the frame is 30% of the area of the picture. How wide is the frame?
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- 9.1 Quadratics – Solving with Radicals
- 9.2 Quadratics – Solving with Exponents
- 9.3 Quadratics – Complete the Square
- 9.4 Quadratics – Quadratic Formula
- 9.5 Quadratics – Build Quadratics From Roots
- 9.6 Quadratics – Quadratic in Form
- 9.7 Quadratics – Rectangles
