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Difficulty: Very hardSolving Quadratic Equations by Factoring

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

  • 3x(4x+5)=5−2x3x(4x + 5) = 5 - 2xx∈{−53,14}x \in \{-\frac{5}{3}, \frac{1}{4}\}
  • x(12x+1)=35x(12x + 1) = 35x∈{−74,53}x \in \{-\frac{7}{4}, \frac{5}{3}\}
  • (2x−3)(3x+1)=5(1−x)(2x - 3)(3x + 1) = 5(1 - x)x∈{−1,43}x \in \{-1, \frac{4}{3}\}

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.
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Topics

Solving Quadratic Equations by Factoring

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