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Difficulty: HardSolving Quadratic Equations by Factoring

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

  • 2x(x+1)=12−3x2x(x + 1) = 12 - 3x{−4,32}\{-4, \frac{3}{2}\}
  • 3x(x−1)=2(x+4)3x(x - 1) = 2(x + 4){−1,83}\{-1, \frac{8}{3}\}
  • 4x(x−2)=54x(x - 2) = 5{−12,52}\{-\frac{1}{2}, \frac{5}{2}\}

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.
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Topics

Solving Quadratic Equations by Factoring

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