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Solving Quadratic Equations by Factoring

26 questions

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Question 1Question →

Match each of the following quadratic equations to its correct set of real solutions.

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Items

2x(x+1)=12−3x2x(x + 1) = 12 - 3x
3x(x−1)=2(x+4)3x(x - 1) = 2(x + 4)
4x(x−2)=54x(x - 2) = 5

Matches

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Answer

The equation 2x(x+1)=12−3x2x(x + 1) = 12 - 3x matches with the solution set {−4,32}\{-4, \frac{3}{2}\}; the equation 3x(x−1)=2(x+4)3x(x - 1) = 2(x + 4) matches with the solution set {−1,83}\{-1, \frac{8}{3}\}; and the equation 4x(x−2)=54x(x - 2) = 5 matches with the solution set {−12,52}\{-\frac{1}{2}, \frac{5}{2}\}.
Each equation is correctly solved by first distributing, moving all terms to one side to set the equation to zero, factoring the resulting trinomial over the integers, and then applying the zero product property to find the corresponding solution set.

Step-by-Step Solution

1
Rearrange the first equation 2x(x+1)=12−3x2x(x + 1) = 12 - 3x into standard form ax2+bx+c=0ax^2 + bx + c = 0.
2x2+5x−12=02x^2 + 5x - 12 = 0
Distributing the 2x2x gives 2x2+2x=12−3x2x^2 + 2x = 12 - 3x. Adding 3x3x and subtracting 1212 from both sides moves all terms to one side.
2
Factor the rearranged first equation 2x2+5x−12=02x^2 + 5x - 12 = 0 and solve for xx.
x=−4x = -4 or x=32x = \frac{3}{2}
Finding two integers that multiply to −24-24 and add to 55 gives 88 and −3-3. Splitting the middle term and factoring by grouping yields (2x−3)(x+4)=0(2x - 3)(x + 4) = 0. Setting each factor to zero gives the solutions.
3
Rearrange the second equation 3x(x−1)=2(x+4)3x(x - 1) = 2(x + 4) into standard form ax2+bx+c=0ax^2 + bx + c = 0.
3x2−5x−8=03x^2 - 5x - 8 = 0
Distributing on both sides gives 3x2−3x=2x+83x^2 - 3x = 2x + 8. Subtracting 2x2x and 88 from both sides sets the quadratic expression to zero.
4
Factor the rearranged second equation 3x2−5x−8=03x^2 - 5x - 8 = 0 and solve for xx.
x=−1x = -1 or x=83x = \frac{8}{3}
Finding two integers that multiply to −24-24 and add to −5-5 gives −8-8 and 33. Grouping terms gives (3x−8)(x+1)=0(3x - 8)(x + 1) = 0. Setting the factors to zero gives the solutions.
5
Rearrange the third equation 4x(x−2)=54x(x - 2) = 5 into standard form ax2+bx+c=0ax^2 + bx + c = 0.
4x2−8x−5=04x^2 - 8x - 5 = 0
Distributing the 4x4x yields 4x2−8x=54x^2 - 8x = 5. Subtracting 55 from both sides sets the equation to zero.
6
Factor the rearranged third equation 4x2−8x−5=04x^2 - 8x - 5 = 0 and solve for xx.
x=−12x = -\frac{1}{2} or x=52x = \frac{5}{2}
Finding two integers that multiply to −20-20 and add to −8-8 gives −10-10 and 22. Grouping terms gives (2x+1)(2x−5)=0(2x + 1)(2x - 5) = 0. Solving each linear factor for xx provides the solutions.

Key Concept

Solving quadratic equations by rearranging them into standard form, factoring by grouping, and applying the zero product property.
Question 2Question →

If 3x(x−3)=2(x+3)−123x(x - 3) = 2(x + 3) - 12, what is the product of the two solutions to this equation?

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Answer: 2

Answer

The product of the two solutions to the equation is 2.
The correct answer is 2. Expanding both sides of the equation 3x(x−3)=2(x+3)−123x(x - 3) = 2(x + 3) - 12 gives 3x2−9x=2x+6−123x^2 - 9x = 2x + 6 - 12, which simplifies to 3x2−9x=2x−63x^2 - 9x = 2x - 6. Subtracting 2x−62x - 6 from both sides results in the standard form quadratic equation 3x2−11x+6=03x^2 - 11x + 6 = 0. Factoring this expression gives (3x−2)(x−3)=0(3x - 2)(x - 3) = 0. Setting each factor to zero yields the solutions x=23x = \frac{2}{3} and x=3x = 3. Multiplying these solutions gives 23×3=2\frac{2}{3} \times 3 = 2. Alternatively, by Vieta's formulas, the product of the roots of a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 is ca\frac{c}{a}, which directly gives 63=2\frac{6}{3} = 2.

Step-by-Step Solution

1
Expand the expressions on both sides of the equation.
3x2−9x=2x+6−123x^2 - 9x = 2x + 6 - 12, which simplifies to 3x2−9x=2x−63x^2 - 9x = 2x - 6
To prepare the equation for rearrangement into the standard quadratic form.
2
Move all terms to the left side of the equation to set it equal to zero.
3x2−11x+6=03x^2 - 11x + 6 = 0
To write the quadratic equation in the standard form ax2+bx+c=0ax^2 + bx + c = 0.
3
Factor the quadratic trinomial by grouping.
(3x−2)(x−3)=0(3x - 2)(x - 3) = 0
To break down the quadratic equation into linear factors that can be solved individually.
4
Set each linear factor to zero and solve for xx.
x=23x = \frac{2}{3} and x=3x = 3
According to the zero product property, if a product of factors is zero, at least one factor must be zero.
5
Calculate the product of the two solutions.
23×3=2\frac{2}{3} \times 3 = 2
To find the product of the solutions as requested by the question.

Key Concept

Solving quadratic equations by factoring after expanding and rearranging terms
Question 3Question →

What is the positive difference between the two real solutions to the quadratic equation 2x(3x−5)=3(x−2)2x(3x - 5) = 3(x - 2)?

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Answer: 56\frac{5}{6}

Answer

The positive difference between the two solutions is 56\frac{5}{6}.
The correct positive difference is 56\frac{5}{6} because rearranging the equation 2x(3x−5)=3(x−2)2x(3x - 5) = 3(x - 2) yields 6x2−13x+6=06x^2 - 13x + 6 = 0. Factoring this equation gives (2x−3)(3x−2)=0(2x - 3)(3x - 2) = 0, which has the solutions x=32x = \frac{3}{2} and x=23x = \frac{2}{3}. Subtracting these values gives 32−23=9−46=56\frac{3}{2} - \frac{2}{3} = \frac{9 - 4}{6} = \frac{5}{6}.

Step-by-Step Solution

1
Expand both sides of the equation.
6x2−10x=3x−66x^2 - 10x = 3x - 6
To begin solving, distribute the term 2x2x on the left side and the constant 33 on the right side.
2
Rearrange the terms to set the quadratic equation to zero.
6x2−13x+6=06x^2 - 13x + 6 = 0
Subtract 3x3x and add 66 to both sides of the equation to write it in standard form ax2+bx+c=0ax^2 + bx + c = 0.
3
Factor the quadratic equation over the integers.
(2x−3)(3x−2)=0(2x - 3)(3x - 2) = 0
Find two binomials whose product is 6x2−13x+66x^2 - 13x + 6. We search for two numbers that multiply to 3636 (from 6×66 \times 6) and sum to −13-13, which are −9-9 and −4-4, allowing factoring by grouping.
4
Solve for the roots of the equation.
x=32x = \frac{3}{2} or x=23x = \frac{2}{3}
Set each linear factor equal to zero using the Zero Product Property: 2x−3=0  ⟹  x=322x - 3 = 0 \implies x = \frac{3}{2} and 3x−2=0  ⟹  x=233x - 2 = 0 \implies x = \frac{2}{3}.
5
Calculate the positive difference between the two solutions.
32−23=96−46=56\frac{3}{2} - \frac{2}{3} = \frac{9}{6} - \frac{4}{6} = \frac{5}{6}
Subtract the smaller solution from the larger solution to find the positive difference.

Key Concept

Solving quadratic equations by rearranging and factoring over integers.
Question 4Question →

The square of 33 less than a number xx is equal to 1616. What is the greater of the two possible values for xx?

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Answer: 7

Answer

7
The word problem translates to the equation (x−3)2=16(x-3)^2 = 16. Expanding the left side and subtracting 16 from both sides gives the standard quadratic equation x2−6x−7=0x^2 - 6x - 7 = 0. This factors into (x−7)(x+1)=0(x-7)(x+1) = 0, yielding two solutions: x=7x = 7 and x=−1x = -1. The greater of these two values is 7.

Step-by-Step Solution

1
Translate the verbal description into an algebraic equation.
(x−3)2=16(x-3)^2 = 16
'3 less than a number xx' is written as x−3x - 3, and its square is set equal to 16.
2
Expand the squared binomial and rearrange the terms into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x2−6x−7=0x^2 - 6x - 7 = 0
Expanding (x−3)2(x-3)^2 yields x2−6x+9x^2 - 6x + 9. Subtracting 16 from both sides sets the equation to zero.
3
Factor the quadratic equation over the integers.
(x−7)(x+1)=0(x - 7)(x + 1) = 0
We find two integers that multiply to −7-7 and add to −6-6, which are −7-7 and 11.
4
Solve for xx by setting each factor to zero, then select the greater value.
x=7x = 7 (since the solutions are 77 and −1-1)
Setting the factors to zero gives x−7=0  ⟹  x=7x - 7 = 0 \implies x = 7 and x+1=0  ⟹  x=−1x + 1 = 0 \implies x = -1. The larger of these two values is 77.

Key Concept

Solving Quadratic Equations by Factoring
Estimated Time:1m 0s
Question 5Question →

The square of 3 less than twice a certain real number is equal to 3 less than 7 times that number. If pp and qq are the two distinct real solutions to this equation with p>qp > q, what is the value of 4p−4q4p - 4q?

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Answer: 13

Answer

13
Correctly expanding the squared expression leads to the quadratic equation 4x2−19x+12=04x^2 - 19x + 12 = 0. Factoring this equation yields the solutions 44 and 3/43/4. Since the problem specifies p>qp > q, we have p=4p = 4 and q=3/4q = 3/4. Substituting these values into the expression 4p−4q4p - 4q gives 4(4)−4(3/4)=16−3=134(4) - 4(3/4) = 16 - 3 = 13.

Step-by-Step Solution

1
Translate the verbal description into an algebraic equation.
(2x−3)2=7x−3(2x - 3)^2 = 7x - 3
'Twice a number' is represented as 2x2x, '3 less than twice the number' is 2x−32x - 3, and 'the square of' that is (2x−3)2(2x - 3)^2. This is set equal to '3 less than 7 times that number', which is 7x−37x - 3.
2
Expand the binomial and write the equation in standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
4x2−19x+12=04x^2 - 19x + 12 = 0
Expanding the left side yields 4x2−12x+9=7x−34x^2 - 12x + 9 = 7x - 3. Subtracting 7x7x and adding 33 to both sides results in the standard form.
3
Factor the quadratic equation by grouping.
(4x−3)(x−4)=0(4x - 3)(x - 4) = 0
We need two numbers that multiply to 4×12=484 \times 12 = 48 and add to −19-19. These numbers are −16-16 and −3-3. Rewriting the equation as 4x2−16x−3x+12=04x^2 - 16x - 3x + 12 = 0 allows us to factor out 4x(x−4)−3(x−4)=04x(x - 4) - 3(x - 4) = 0, which simplifies to (4x−3)(x−4)=0(4x - 3)(x - 4) = 0.
4
Find the roots of the equation and evaluate the target expression 4p−4q4p - 4q.
p=4p = 4, q=3/4q = 3/4, and the final value is 1313.
Setting each factor to zero gives x=3/4x = 3/4 and x=4x = 4. Since p>qp > q, we assign p=4p = 4 and q=3/4q = 3/4. Evaluating the expression yields 4(4)−4(3/4)=16−3=134(4) - 4(3/4) = 16 - 3 = 13.

Key Concept

Solving Quadratic Equations by Factoring
Estimated Time:2m 30s
Question 6Question →

The sum of 99 and the product of a number and 66 less than that number is equal to 33 more than the number. What is the sum of all possible values of this number?

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Answer: 7

Answer

7
The correct answer is 77. The verbal description translates to x(x−6)+9=x+3x(x - 6) + 9 = x + 3. Distributing the left side gives x2−6x+9=x+3x^2 - 6x + 9 = x + 3. Subtracting xx and 33 from both sides results in x2−7x+6=0x^2 - 7x + 6 = 0. Factoring the quadratic expression gives (x−6)(x−1)=0(x - 6)(x - 1) = 0, which yields the solutions x=6x = 6 and x=1x = 1. Summing these values gives 6+1=76 + 1 = 7.

Step-by-Step Solution

1
Translate the verbal statement into an algebraic equation.
x(x−6)+9=x+3x(x - 6) + 9 = x + 3, where xx represents the unknown number.
Establishing the relationship between the algebraic expressions defined by the problem.
2
Expand the left side of the equation and combine like terms to set the equation to zero.
x2−6x+9=x+3  ⟹  x2−7x+6=0x^2 - 6x + 9 = x + 3 \implies x^2 - 7x + 6 = 0.
Quadratic equations must be set to zero before they can be solved by factoring.
3
Factor the quadratic equation over the integers.
(x−6)(x−1)=0(x - 6)(x - 1) = 0.
Finding two numbers that multiply to 66 and add up to −7-7 allows us to factor the trinomial.
4
Apply the zero-product property to find the individual roots.
x=6x = 6 and x=1x = 1.
If the product of two factors is zero, at least one of the factors must be zero.
5
Sum the possible values of the number.
6+1=76 + 1 = 7.
The question asks for the sum of all possible values of the number.

Key Concept

Solving Quadratic Equations by Factoring
Question 7Question →

Solve each quadratic equation by factoring, and match the equation to its correct solution set.

Click a left item, then click its matching right item

Items

3x(4x+5)=5−2x3x(4x + 5) = 5 - 2x
x(12x+1)=35x(12x + 1) = 35
(2x−3)(3x+1)=5(1−x)(2x - 3)(3x + 1) = 5(1 - x)

Matches

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Answer

The equation 3x(4x+5)=5−2x3x(4x + 5) = 5 - 2x matches the solution set {−53,14}\{-\frac{5}{3}, \frac{1}{4}\}; the equation x(12x+1)=35x(12x + 1) = 35 matches the solution set {−74,53}\{-\frac{7}{4}, \frac{5}{3}\}; and the equation (2x−3)(3x+1)=5(1−x)(2x - 3)(3x + 1) = 5(1 - x) matches the solution set {−1,43}\{-1, \frac{4}{3}\}.
Each of the quadratic equations is solved by first expanding any products, collecting all terms on the left-hand side to establish the standard form ax2+bx+c=0ax^2 + bx + c = 0, dividing by any common numerical factors, factoring the resulting quadratic expression into two linear binomials, and solving each linear equation for xx. This correctly pairs the first equation with {−53,14}\{-\frac{5}{3}, \frac{1}{4}\}, the second equation with {−74,53}\{-\frac{7}{4}, \frac{5}{3}\}, and the third equation with {−1,43}\{-1, \frac{4}{3}\}.

Step-by-Step Solution

1
Solve 3x(4x+5)=5−2x3x(4x + 5) = 5 - 2x.
12x2+17x−5=0⇒(3x+5)(4x−1)=0⇒x=−5312x^2 + 17x - 5 = 0 \Rightarrow (3x + 5)(4x - 1) = 0 \Rightarrow x = -\frac{5}{3} or x=14x = \frac{1}{4}.
Distribute the term on the left, rearrange the terms to set the equation to zero, find factors of 12×(−5)=−6012 \times (-5) = -60 that sum to 1717 (which are 2020 and −3-3), factor by grouping, and apply the Zero Product Property.
2
Solve x(12x+1)=35x(12x + 1) = 35.
12x2+x−35=0⇒(3x−5)(4x+7)=0⇒x=5312x^2 + x - 35 = 0 \Rightarrow (3x - 5)(4x + 7) = 0 \Rightarrow x = \frac{5}{3} or x=−74x = -\frac{7}{4}.
Expand the left side, subtract 3535 from both sides, find factors of 12×(−35)=−42012 \times (-35) = -420 that sum to 11 (which are 2121 and −20-20), factor by grouping, and solve for xx.
3
Solve (2x−3)(3x+1)=5(1−x)(2x - 3)(3x + 1) = 5(1 - x).
6x2−2x−8=0⇒3x2−x−4=0⇒(3x−4)(x+1)=0⇒x=436x^2 - 2x - 8 = 0 \Rightarrow 3x^2 - x - 4 = 0 \Rightarrow (3x - 4)(x + 1) = 0 \Rightarrow x = \frac{4}{3} or x=−1x = -1.
Expand both sides, move all terms to the left, divide the quadratic equation by 22 to simplify, factor the trinomial, and solve for the roots.

Key Concept

Rearranging non-standard quadratic equations into the standard form ax2+bx+c=0ax^2 + bx + c = 0 and solving them by factoring over the integers.
Question 8Question →

Match each quadratic equation with its correct solution set by solving the equation by factoring.

Click a left item, then click its matching right item

Items

x(x−1)=12x(x - 1) = 12
2x2+5x=32x^2 + 5x = 3
3x2+8=10x3x^2 + 8 = 10x
2x2−24=8x2x^2 - 24 = 8x

Matches

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Answer

The equation x(x−1)=12x(x - 1) = 12 matches the solution set {−3,4}\{-3, 4\}; the equation 2x2+5x=32x^2 + 5x = 3 matches the solution set {−3,12}\{-3, \frac{1}{2}\}; the equation 3x2+8=10x3x^2 + 8 = 10x matches the solution set {43,2}\{\frac{4}{3}, 2\}; and the equation 2x2−24=8x2x^2 - 24 = 8x matches the solution set {−2,6}\{-2, 6\}.
Each quadratic equation is correctly matched to its solutions by first rewriting the equation in standard form ax2+bx+c=0ax^2 + bx + c = 0, factoring the trinomial over the integers, and then applying the zero product property to find the roots.

Step-by-Step Solution

1
Set each quadratic equation to standard form ax2+bx+c=0ax^2 + bx + c = 0 by expanding terms and moving all terms to one side.
The equations become:
1) x2−x−12=0x^2 - x - 12 = 0
2) 2x2+5x−3=02x^2 + 5x - 3 = 0
3) 3x2−10x+8=03x^2 - 10x + 8 = 0
4) 2x2−8x−24=02x^2 - 8x - 24 = 0
Before a quadratic equation can be solved by factoring, it must be set equal to zero so that the zero product property can be applied.
2
Factor each quadratic expression completely over the integers.
The factored expressions are:
1) (x−4)(x+3)=0(x - 4)(x + 3) = 0
2) (2x−1)(x+3)=0(2x - 1)(x + 3) = 0
3) (3x−4)(x−2)=0(3x - 4)(x - 2) = 0
4) 2(x−6)(x+2)=02(x - 6)(x + 2) = 0
Factoring rewrites the quadratic expressions as products of linear factors.
3
Apply the zero product property by setting each linear factor equal to zero and solving for xx.
The solution sets are:
1) x=4x = 4 or x=−3x = -3, yielding {−3,4}\{-3, 4\}
2) x=12x = \frac{1}{2} or x=−3x = -3, yielding {−3,12}\{-3, \frac{1}{2}\}
3) x=43x = \frac{4}{3} or x=2x = 2, yielding {43,2}\{\frac{4}{3}, 2\}
4) x=6x = 6 or x=−2x = -2, yielding {−2,6}\{-2, 6\}
If the product of two or more algebraic factors is zero, then at least one of the individual factors must equal zero.

Key Concept

Solving Quadratic Equations by Factoring

Alternative Method

You can verify the solution sets by substituting the values of the roots back into the original equations to check if they yield a true statement, or by using the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} as an alternative algebraic method.
Estimated Time:1m 30s
Question 9Question →

Match each quadratic equation with its correct set of real solutions.

Click a left item, then click its matching right item

Items

x2−5x+6=0x^2 - 5x + 6 = 0
x2+5x+6=0x^2 + 5x + 6 = 0
x2−x−6=0x^2 - x - 6 = 0

Matches

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Answer

The equation x2−5x+6=0x^2 - 5x + 6 = 0 matches with the solutions x=2x = 2 and x=3x = 3. The equation x2+5x+6=0x^2 + 5x + 6 = 0 matches with the solutions x=−3x = -3 and x=−2x = -2. The equation x2−x−6=0x^2 - x - 6 = 0 matches with the solutions x=−2x = -2 and x=3x = 3.
Each equation is solved by factoring the quadratic trinomial into two binomials, then applying the zero product property to find the values of xx that make each factor zero. Specifically, x2−5x+6=0x^2 - 5x + 6 = 0 factors into (x−2)(x−3)=0(x - 2)(x - 3) = 0, yielding solutions x=2x = 2 and x=3x = 3. The equation x2+5x+6=0x^2 + 5x + 6 = 0 factors into (x+2)(x+3)=0(x + 2)(x + 3) = 0, yielding solutions x=−2x = -2 and x=−3x = -3. Finally, x2−x−6=0x^2 - x - 6 = 0 factors into (x−3)(x+2)=0(x - 3)(x + 2) = 0, yielding solutions x=3x = 3 and x=−2x = -2.

Step-by-Step Solution

1
Factor the quadratic equation x2−5x+6=0x^2 - 5x + 6 = 0.
(x−2)(x−3)=0(x - 2)(x - 3) = 0
Identify two integers that multiply to 66 and add up to −5-5. These integers are −2-2 and −3-3.
2
Solve for xx by setting each linear factor equal to zero: x−2=0x - 2 = 0 and x−3=0x - 3 = 0.
x=2x = 2 and x=3x = 3
Applying the zero product property means if the product of two numbers is zero, at least one of them must be zero.
3
Factor the quadratic equation x2+5x+6=0x^2 + 5x + 6 = 0.
(x+2)(x+3)=0(x + 2)(x + 3) = 0
Identify two integers that multiply to 66 and add up to 55. These integers are 22 and 33.
4
Solve for xx by setting each linear factor equal to zero: x+2=0x + 2 = 0 and x+3=0x + 3 = 0.
x=−2x = -2 and x=−3x = -3
Applying the zero product property gives the solutions as the negations of the terms inside the binomials.
5
Factor the quadratic equation x2−x−6=0x^2 - x - 6 = 0.
(x−3)(x+2)=0(x - 3)(x + 2) = 0
Identify two integers that multiply to −6-6 and add up to −1-1. These integers are −3-3 and 22.
6
Solve for xx by setting each linear factor equal to zero: x−3=0x - 3 = 0 and x+2=0x + 2 = 0.
x=3x = 3 and x=−2x = -2
Setting the linear factors to zero yields the roots of the equation.

Key Concept

Solving Quadratic Equations by Factoring
Estimated Time:1m 30s
Question 10Question →

What is the sum of the solutions to the quadratic equation x(x−6)=16x(x - 6) = 16?

Show answer & explanation

Answer: 6

Answer

The sum of the solutions is 6.
Rearranging the equation to standard form gives x2−6x−16=0x^2 - 6x - 16 = 0. Factoring the trinomial yields (x−8)(x+2)=0(x - 8)(x + 2) = 0. Solving for xx by setting each factor to zero gives x=8x = 8 and x=−2x = -2. The sum of these solutions is 8+(−2)=68 + (-2) = 6.

Step-by-Step Solution

1
Distribute the variable on the left side of the equation.
x2−6x=16x^2 - 6x = 16
To solve a quadratic equation, we must first expand all products to identify the quadratic terms.
2
Subtract 16 from both sides of the equation to write it in standard form.
x2−6x−16=0x^2 - 6x - 16 = 0
A quadratic equation must be set to zero before factoring.
3
Factor the quadratic trinomial.
(x−8)(x+2)=0(x - 8)(x + 2) = 0
We need to find two numbers that multiply to −16-16 and add to −6-6. These numbers are −8-8 and 22.
4
Set each factor to zero and solve for xx.
x=8x = 8 and x=−2x = -2
According to the zero product property, if the product of two factors is zero, at least one of the factors must be zero.
5
Add the two solutions to find their sum.
8+(−2)=68 + (-2) = 6
The question asks for the sum of the solutions.

Key Concept

Solving quadratic equations by factoring after rearranging terms into standard form.
Question 11Question →

What is the sum of the distinct real solutions to the equation (2x−1)2=(x+2)2(2x - 1)^2 = (x + 2)^2?

Show answer & explanation

Answer: 83\frac{8}{3}

Answer

The sum of the distinct real solutions is 83\frac{8}{3}.
To solve (2x−1)2=(x+2)2(2x - 1)^2 = (x + 2)^2, we rearrange the equation to (2x−1)2−(x+2)2=0(2x - 1)^2 - (x + 2)^2 = 0. Using the difference of squares identity, we factor this into [(2x−1)−(x+2)][(2x−1)+(x+2)]=0[(2x - 1) - (x + 2)][(2x - 1) + (x + 2)] = 0, which simplifies to (x−3)(3x+1)=0(x - 3)(3x + 1) = 0. The solutions are x=3x = 3 and x=−13x = -\frac{1}{3}. Summing these gives 3+(−13)=833 + (-\frac{1}{3}) = \frac{8}{3}.

Step-by-Step Solution

1
Rearrange the equation by moving all terms to one side to set it to zero.
(2x−1)2−(x+2)2=0(2x - 1)^2 - (x + 2)^2 = 0
To solve a quadratic equation by factoring, it must first be set equal to zero.
2
Factor the expression using the difference of squares formula, a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b), where a=2x−1a = 2x - 1 and b=x+2b = x + 2.
[(2x−1)−(x+2)][(2x−1)+(x+2)]=0[(2x - 1) - (x + 2)][(2x - 1) + (x + 2)] = 0
Using the difference of squares allows us to factor the quadratic expression directly without fully expanding it.
3
Simplify the terms inside each set of brackets.
(2x−1−x−2)(2x−1+x+2)=0⇒(x−3)(3x+1)=0(2x - 1 - x - 2)(2x - 1 + x + 2) = 0 \Rightarrow (x - 3)(3x + 1) = 0
Simplifying the binomials reveals the two linear factors of the quadratic equation.
4
Set each linear factor to zero to find the distinct real solutions.
x−3=0⇒x=3x - 3 = 0 \Rightarrow x = 3 and 3x+1=0⇒x=−133x + 1 = 0 \Rightarrow x = -\frac{1}{3}
According to the zero product property, if the product of two factors is zero, at least one of the factors must be zero.
5
Calculate the sum of the distinct real solutions.
3+(−13)=93−13=833 + \left(-\frac{1}{3}\right) = \frac{9}{3} - \frac{1}{3} = \frac{8}{3}
The question asks for the sum of the solutions.

Key Concept

Solving quadratic equations by factoring, specifically utilizing the difference of squares method after rearranging terms.
Question 12Question →

A rectangular garden is surrounded by a uniform gravel path that is 11 foot wide. The length of the garden is 33 feet less than twice its width. If the total area of the garden and the path combined is 117117 square feet, what is the width of the garden, in feet?

Show answer & explanation

Answer: 7

Answer

The width of the garden is 7 feet.
By representing the garden's width as ww, the length is 2w−32w - 3. The combined dimensions including the 1-foot uniform path on all sides are w+2w + 2 and 2w−12w - 1. Setting their product equal to the combined area of 117 square feet gives (w+2)(2w−1)=117(w+2)(2w-1) = 117, which simplifies to the quadratic equation 2w2+3w−119=02w^2 + 3w - 119 = 0. Factoring this equation yields (2w+17)(w−7)=0(2w + 17)(w - 7) = 0, giving the solutions w=−8.5w = -8.5 and w=7w = 7. Since width must be positive, the width of the garden is 77 feet.

Step-by-Step Solution

1
Define variables for the garden's dimensions and the combined dimensions including the path.
Garden width = ww, garden length = 2w−32w - 3. Combined width = w+2w + 2, combined length = 2w−12w - 1.
The path surrounds the garden uniformly, adding 11 foot of width to each of the four sides (adding 22 feet total to both overall width and overall length).
2
Write the area equation for the combined area.
(w+2)(2w−1)=117(w + 2)(2w - 1) = 117
The total area of the garden and path combined is given as 117117 square feet.
3
Expand and rearrange the equation into standard quadratic form aw2+bw+c=0aw^2 + bw + c = 0.
2w2+3w−119=02w^2 + 3w - 119 = 0
Expanding (w+2)(2w−1)(w + 2)(2w - 1) gives 2w2+3w−22w^2 + 3w - 2. Subtracting 117117 from both sides yields the standard form.
4
Factor the quadratic equation over the integers.
(2w+17)(w−7)=0(2w + 17)(w - 7) = 0
We find two numbers that multiply to 2×(−119)=−2382 \times (-119) = -238 and sum to 33. These numbers are 1717 and −14-14. Rewriting the middle term and factoring by grouping yields (2w+17)(w−7)=0(2w + 17)(w - 7) = 0.
5
Solve for ww and select the mathematically and physically valid solution.
w=7w = 7 (discarding the negative root w=−8.5w = -8.5)
A physical measurement like width must be positive.

Key Concept

Solving quadratic word problems by setting up a quadratic equation and solving it by factoring.
Estimated Time:2m 0s
Question 13Question →

The quadratic equation x2−4x−12=0x^2 - 4x - 12 = 0 has two real solutions. What is the value of the positive solution to this equation?

Show answer & explanation

Answer: 6

Answer

The positive solution to the equation is 66.
Factoring the quadratic trinomial x2−4x−12=0x^2 - 4x - 12 = 0 yields (x−6)(x+2)=0(x - 6)(x + 2) = 0. Setting the individual binomial factors to zero gives the solutions x=6x = 6 and x=−2x = -2. The positive solution among these is 66.

Step-by-Step Solution

1
Factor the quadratic equation
(x−6)(x+2)=0(x - 6)(x + 2) = 0
Factoring the trinomial x2−4x−12x^2 - 4x - 12 requires finding two integers whose product is −12-12 and whose sum is −4-4. These numbers are −6-6 and 22.
2
Apply the zero product property
x−6=0x - 6 = 0 or x+2=0x + 2 = 0
If the product of two factors is equal to zero, then at least one of the individual factors must equal zero.
3
Solve for the variable and identify the positive root
x=6x = 6 and x=−2x = -2
Solving the linear equations yields x=6x = 6 and x=−2x = -2. Since the question asks for the positive solution, we select 66.

Key Concept

Solving quadratic equations by factoring
Question 14Question →

A certain real number xx satisfies the condition that the square of 33 less than twice xx is equal to 88 times the quantity 33 minus xx. What is the ratio of the larger solution to the smaller solution of this equation?

Show answer & explanation

Answer: −53-\frac{5}{3}

Answer

The ratio of the larger solution to the smaller solution is −53-\frac{5}{3}.
The correct answer is −53-\frac{5}{3}. The verbal statement translates directly to (2x−3)2=8(3−x)(2x - 3)^2 = 8(3 - x). Expanding both sides yields 4x2−12x+9=24−8x4x^2 - 12x + 9 = 24 - 8x. Rearranging into standard form gives 4x2−4x−15=04x^2 - 4x - 15 = 0. Factoring by grouping yields (2x+3)(2x−5)=0(2x + 3)(2x - 5) = 0, giving the solutions x=−32x = -\frac{3}{2} and x=52x = \frac{5}{2}. The ratio of the larger root to the smaller root is 5/2−3/2=−53\frac{5/2}{-3/2} = -\frac{5}{3}.

Step-by-Step Solution

1
Translate the verbal description into an algebraic equation.
(2x−3)2=8(3−x)(2x - 3)^2 = 8(3 - x)
'Twice xx' is 2x2x, '3 less than twice xx' is 2x−32x - 3, and its square is (2x−3)2(2x - 3)^2. This is equal to 8 times the quantity 3−x3 - x.
2
Expand both sides of the equation.
4x2−12x+9=24−8x4x^2 - 12x + 9 = 24 - 8x
Expanding the binomial (2x−3)2(2x - 3)^2 gives 4x2−12x+94x^2 - 12x + 9 and distributing the right side gives 24−8x24 - 8x.
3
Rearrange the equation to set it equal to zero.
4x2−4x−15=04x^2 - 4x - 15 = 0
Add 8x8x and subtract 2424 from both sides to gather all terms on one side of the equation.
4
Factor the quadratic equation by grouping.
(2x+3)(2x−5)=0(2x + 3)(2x - 5) = 0
Find two numbers that multiply to 4×(−15)=−604 \times (-15) = -60 and add to −4-4. These numbers are −10-10 and 66. Rewrite the middle term as −10x+6x-10x + 6x and factor: 2x(2x−5)+3(2x−5)=0  ⟹  (2x+3)(2x−5)=02x(2x - 5) + 3(2x - 5) = 0 \implies (2x + 3)(2x - 5) = 0.
5
Solve for the roots of the equation.
x=−32x = -\frac{3}{2} and x=52x = \frac{5}{2}
Set each factor equal to zero: 2x+3=0  ⟹  x=−322x + 3 = 0 \implies x = -\frac{3}{2} and 2x−5=0  ⟹  x=522x - 5 = 0 \implies x = \frac{5}{2}.
6
Identify the larger and smaller solutions and compute their ratio.
−53-\frac{5}{3}
The larger solution is 52\frac{5}{2} and the smaller solution is −32-\frac{3}{2}. Their ratio is 5/2−3/2=−53\frac{5/2}{-3/2} = -\frac{5}{3}.

Key Concept

Solving quadratic equations of the form ax2+bx+c=0ax^2 + bx + c = 0 by factoring over the integers.
Question 15Question →

What is the sum of all real values of xx that satisfy the equation (x−3)2+x(x+2)=15(x - 3)^2 + x(x + 2) = 15?

Show answer & explanation

Answer: 2

Answer

The sum of all real values of xx that satisfy the equation is 22.
Expanding the equation yields 2x2−4x+9=152x^2 - 4x + 9 = 15. Setting this to zero gives 2x2−4x−6=02x^2 - 4x - 6 = 0. Dividing by the common factor of 22 simplifies this to x2−2x−3=0x^2 - 2x - 3 = 0. Factoring the trinomial yields (x−3)(x+1)=0(x - 3)(x + 1) = 0, which gives the two solutions x=3x = 3 and x=−1x = -1. Summing these two solutions gives 3+(−1)=23 + (-1) = 2.

Step-by-Step Solution

1
Expand both terms on the left side of the equation.
(x2−6x+9)+(x2+2x)=15(x^2 - 6x + 9) + (x^2 + 2x) = 15
Applying the binomial squaring formula (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2 to (x−3)2(x - 3)^2 and distributing xx to both terms in x(x+2)x(x + 2) allows us to simplify the equation.
2
Combine like terms and set the quadratic equation to zero.
2x2−4x−6=02x^2 - 4x - 6 = 0
Grouping x2x^2 terms, xx terms, and constant terms on one side is necessary to format the quadratic equation as ax2+bx+c=0ax^2 + bx + c = 0 before factoring.
3
Divide the entire equation by the common factor of 22 to simplify factoring.
x2−2x−3=0x^2 - 2x - 3 = 0
Simplifying the quadratic equation makes it easier to find two binomial factors.
4
Factor the quadratic trinomial by finding two numbers that multiply to −3-3 and add to −2-2.
(x−3)(x+1)=0(x - 3)(x + 1) = 0
Since −3×1=−3-3 \times 1 = -3 and −3+1=−2-3 + 1 = -2, we can write the quadratic in factored form.
5
Set each factor to zero to solve for xx.
x=3x = 3 or x=−1x = -1
Applying the zero product property determines the two values of xx that satisfy the original equation.
6
Calculate the sum of the two solutions.
3+(−1)=23 + (-1) = 2
The question asks for the sum of all real values of xx that satisfy the equation.

Key Concept

Solving quadratic equations by rearranging terms, factoring trinomials, and applying the zero product property.
Question 16Question →

For a certain positive number yy, the product of yy and the quantity 2y+52y + 5 is equal to 1212. What is the value of yy?

Show answer & explanation

Answer: 1.5

Answer

The positive value of yy is 1.51.5.
The correct answer is 1.51.5. By translating the word problem, we obtain y(2y+5)=12y(2y + 5) = 12. Expanding this gives 2y2+5y=122y^2 + 5y = 12, and subtracting 1212 from both sides yields the standard quadratic equation 2y2+5y−12=02y^2 + 5y - 12 = 0. Factoring this expression gives (2y−3)(y+4)=0(2y - 3)(y + 4) = 0. Setting the first factor to zero yields y=1.5y = 1.5, which is positive and therefore satisfies the given condition.

Step-by-Step Solution

1
Write the equation representing the relationship.
y(2y+5)=12y(2y + 5) = 12
To translate the verbal description into an algebraic equation.
2
Distribute yy and set the equation equal to zero.
2y2+5y−12=02y^2 + 5y - 12 = 0
To put the quadratic equation into standard form ay2+by+c=0ay^2 + by + c = 0 so it can be factored.
3
Factor the quadratic equation.
(2y−3)(y+4)=0(2y - 3)(y + 4) = 0
To find the factors that multiply to give the quadratic expression.
4
Solve for yy and apply the constraint.
y=1.5y = 1.5
Setting the factors to zero gives y=1.5y = 1.5 and y=−4y = -4. Since the problem states yy is positive, we select the positive root.

Key Concept

Solving a non-monic quadratic equation by factoring after translating a verbal description into algebra.
Question 17Question →

If x(x−6)=7x(x - 6) = 7, what is the positive difference between the two solutions to this equation?

Show answer & explanation

Answer: 8

Answer

The positive difference between the two solutions is 88.
First, expand the left side of the equation to get x2−6x=7x^2 - 6x = 7. Next, subtract 77 from both sides to write the equation in standard form: x2−6x−7=0x^2 - 6x - 7 = 0. Factoring this quadratic expression gives (x−7)(x+1)=0(x - 7)(x + 1) = 0. Setting each factor to zero yields the solutions x=7x = 7 and x=−1x = -1. The positive difference between these two solutions is 7−(−1)=87 - (-1) = 8.

Step-by-Step Solution

1
Distribute the xx on the left side of the equation.
x2−6x=7x^2 - 6x = 7
Multiplying xx by (x−6)(x - 6) expands the left side to prepare for standard form.
2
Subtract 77 from both sides to set the quadratic equation to zero.
x2−6x−7=0x^2 - 6x - 7 = 0
Setting the quadratic equation to standard form ax2+bx+c=0ax^2 + bx + c = 0 is necessary before factoring.
3
Factor the quadratic trinomial.
(x−7)(x+1)=0(x - 7)(x + 1) = 0
We need two numbers that multiply to −7-7 and add to −6-6, which are −7-7 and 11.
4
Set each factor to zero to find the solutions.
x=7x = 7 and x=−1x = -1
The zero product property states that if a product is zero, at least one factor must be zero.
5
Calculate the positive difference between the two solutions.
7−(−1)=87 - (-1) = 8
The positive difference is found by subtracting the smaller solution from the larger solution.

Key Concept

Solving Quadratic Equations by Factoring
Estimated Time:1m 0s
Question 18Question →

What is the positive difference between the two real solutions to the equation (2x−1)2=x(3x−5)+7(2x - 1)^2 = x(3x - 5) + 7?

Show answer & explanation

Answer: 5

Answer

The positive difference between the two real solutions is 5.
The correct answer is 5. By expanding the equation, we get 4x2−4x+1=3x2−5x+74x^2 - 4x + 1 = 3x^2 - 5x + 7. Rearranging the terms to set the equation to zero yields x2+x−6=0x^2 + x - 6 = 0. Factoring this equation gives (x+3)(x−2)=0(x + 3)(x - 2) = 0, which has the solutions x=−3x = -3 and x=2x = 2. The positive difference between these two solutions is ∣2−(−3)∣=5|2 - (-3)| = 5.

Step-by-Step Solution

1
Expand both sides of the equation.
4x2−4x+1=3x2−5x+74x^2 - 4x + 1 = 3x^2 - 5x + 7
Expanding the squared term on the left side and distributing the xx on the right side allows us to write the equation in polynomial form.
2
Rearrange the equation by moving all terms to the left side to set the right side to zero.
x2+x−6=0x^2 + x - 6 = 0
Subtracting 3x23x^2, adding 5x5x, and subtracting 77 from both sides simplifies the equation into standard quadratic form, ax2+bx+c=0ax^2 + bx + c = 0.
3
Factor the quadratic trinomial.
(x+3)(x−2)=0(x + 3)(x - 2) = 0
Finding two integers that multiply to −6-6 and add to 11 gives 33 and −2-2, allowing us to factor the equation over the integers.
4
Solve for xx by setting each linear factor to zero.
x=−3x = -3 and x=2x = 2
According to the zero-product property, if the product of two factors is zero, at least one of the factors must be zero.
5
Calculate the positive difference between the two solutions.
∣2−(−3)∣=5|2 - (-3)| = 5
Subtracting the smaller solution from the larger solution gives the positive distance between them on the number line.

Key Concept

Solving quadratic equations by expanding, rearranging into standard form, and factoring over the integers.
Estimated Time:2m 0s
Question 19Question →

What is the positive difference between the two real solutions to the equation (x−1)2=5x−5(x - 1)^2 = 5x - 5?

Show answer & explanation

Answer: 5

Answer

The positive difference between the two real solutions is 5.
Expanding the left side of (x−1)2=5x−5(x - 1)^2 = 5x - 5 yields x2−2x+1=5x−5x^2 - 2x + 1 = 5x - 5. Moving all terms to the left side by subtracting 5x5x and adding 55 gives the standard quadratic equation x2−7x+6=0x^2 - 7x + 6 = 0. Factoring this expression gives (x−1)(x−6)=0(x - 1)(x - 6) = 0, which yields the solutions 11 and 66. The positive difference between these solutions is 6−1=56 - 1 = 5.

Step-by-Step Solution

1
Expand the squared binomial on the left side of the equation.
x2−2x+1=5x−5x^2 - 2x + 1 = 5x - 5
Before factoring a quadratic equation, all terms must be expanded and moved to one side to set the equation equal to zero.
2
Subtract 5x5x and add 55 to both sides to rearrange the equation into standard quadratic form.
x2−7x+6=0x^2 - 7x + 6 = 0
This sets the equation equal to zero, which is a prerequisite for using the zero product property.
3
Factor the quadratic expression by finding two numbers that multiply to 66 and add to −7-7.
(x−1)(x−6)=0(x - 1)(x - 6) = 0
Factoring allows us to split the quadratic equation into two linear equations.
4
Set each factor to zero to solve for xx.
x=1x = 1 and x=6x = 6
By the zero product property, if the product of two factors is zero, at least one of the factors must be zero.
5
Subtract the smaller solution from the larger solution to find the positive difference.
6−1=56 - 1 = 5
The question asks for the positive difference between the two solutions.

Key Concept

Solving quadratic equations by expanding, rearranging into standard form, and factoring over the integers.
Question 20Question →

For each of the given quadratic equations, solve for xx by factoring. Match each quadratic equation on the left to its correct solution set on the right.

Click a left item, then click its matching right item

Items

The equation 2x2+5x=32x^2 + 5x = 3
The equation 3x2−10x+8=03x^2 - 10x + 8 = 0
The equation x(x+2)=15x(x + 2) = 15

Matches

Show answer & explanation

Answer

The equation 2x2+5x=32x^2 + 5x = 3 matches the solution set {−3,12}\left\{-3, \frac{1}{2}\right\}; the equation 3x2−10x+8=03x^2 - 10x + 8 = 0 matches the solution set {43,2}\left\{\frac{4}{3}, 2\right\}; and the equation x(x+2)=15x(x + 2) = 15 matches the solution set {−5,3}\left\{-5, 3\right\}.
Each equation matches its corresponding solution set through distributing terms if necessary, rewriting the equation in standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0, factoring the quadratic trinomial over the integers, and then using the zero product property to solve for xx.

Step-by-Step Solution

1
For the equation 2x2+5x=32x^2 + 5x = 3, rewrite in standard form ax2+bx+c=0ax^2 + bx + c = 0 by subtracting 3 from both sides to get 2x2+5x−3=02x^2 + 5x - 3 = 0.
The equation is rewritten as 2x2+5x−3=02x^2 + 5x - 3 = 0.
Before factoring a quadratic equation, all terms must be moved to one side so the other side is equal to zero.
2
Factor the trinomial 2x2+5x−32x^2 + 5x - 3 by grouping. Find two integers that multiply to 2×(−3)=−62 \times (-3) = -6 and add to 55. These integers are 66 and −1-1. Rewrite the middle term and factor by grouping: 2x2+6x−x−3=2x(x+3)−1(x+3)=(2x−1)(x+3)=02x^2 + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3) = 0.
The equation becomes (2x−1)(x+3)=0(2x - 1)(x + 3) = 0.
Factoring allows us to apply the zero product property to find the solutions.
3
Set each factor of (2x−1)(x+3)=0(2x - 1)(x + 3) = 0 to zero and solve for xx: 2x−1=0⇒x=122x - 1 = 0 \Rightarrow x = \frac{1}{2} and x+3=0⇒x=−3x + 3 = 0 \Rightarrow x = -3.
The solutions are x=12x = \frac{1}{2} and x=−3x = -3, forming the solution set {−3,12}\left\{-3, \frac{1}{2}\right\}.
By the zero product property, if the product of two factors is zero, at least one of the factors must be zero.
4
For the equation 3x2−10x+8=03x^2 - 10x + 8 = 0, factor the trinomial by finding two integers that multiply to 3×8=243 \times 8 = 24 and add to −10-10. These integers are −6-6 and −4-4. Rewrite the middle term and factor by grouping: 3x2−6x−4x+8=3x(x−2)−4(x−2)=(3x−4)(x−2)=03x^2 - 6x - 4x + 8 = 3x(x - 2) - 4(x - 2) = (3x - 4)(x - 2) = 0.
The equation becomes (3x−4)(x−2)=0(3x - 4)(x - 2) = 0.
The equation is already in standard form, so we can directly proceed with factoring.
5
Set each factor of (3x−4)(x−2)=0(3x - 4)(x - 2) = 0 to zero and solve for xx: 3x−4=0⇒x=433x - 4 = 0 \Rightarrow x = \frac{4}{3} and x−2=0⇒x=2x - 2 = 0 \Rightarrow x = 2.
The solutions are x=43x = \frac{4}{3} and x=2x = 2, forming the solution set {43,2}\left\{\frac{4}{3}, 2\right\}.
Solving each linear factor yields the roots of the quadratic equation.
6
For the equation x(x+2)=15x(x + 2) = 15, first distribute xx to get x2+2x=15x^2 + 2x = 15, then subtract 15 from both sides to write in standard form: x2+2x−15=0x^2 + 2x - 15 = 0.
The equation is rewritten as x2+2x−15=0x^2 + 2x - 15 = 0.
Distributing and moving terms sets the quadratic to zero, which is necessary for factoring.
7
Factor the quadratic x2+2x−15=0x^2 + 2x - 15 = 0 by finding two integers that multiply to −15-15 and add to 22. These integers are 55 and −3-3, yielding (x+5)(x−3)=0(x + 5)(x - 3) = 0.
The equation becomes (x+5)(x−3)=0(x + 5)(x - 3) = 0.
Factoring a quadratic trinomial with a leading coefficient of 1 involves finding numbers that sum to the linear coefficient and multiply to the constant term.
8
Set each factor of (x+5)(x−3)=0(x + 5)(x - 3) = 0 to zero and solve for xx: x+5=0⇒x=−5x + 5 = 0 \Rightarrow x = -5 and x−3=0⇒x=3x - 3 = 0 \Rightarrow x = 3.
The solutions are x=−5x = -5 and x=3x = 3, forming the solution set {−5,3}\left\{-5, 3\right\}.
Solving the resulting linear equations gives the roots of the original quadratic equation.

Key Concept

Solving Quadratic Equations by Factoring

Alternative Method

Instead of factoring, the solutions to these quadratic equations can be verified by substituting the values in the solution sets back into the original equations, or by applying the quadratic formula: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} after rewriting them in standard form.
Estimated Time:2m 0s
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