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Difficulty: MediumSolving Quadratic Equations by Factoring

If the equation 3x2+5x=23x^2 + 5x = 2 is solved for xx, what is the positive difference between the two solutions?

  1. A
    13\frac{1}{3}
  2. B
    43\frac{4}{3}
  3. C
    53\frac{5}{3}
  4. 73\frac{7}{3}Answer
  5. E
    33

Answer

The positive difference between the two solutions is 73\frac{7}{3}.
Rearranging the equation 3x2+5x=23x^2 + 5x = 2 by subtracting 22 from both sides gives the standard quadratic equation 3x2+5x−2=03x^2 + 5x - 2 = 0. Factoring this expression yields (3x−1)(x+2)=0(3x - 1)(x + 2) = 0. Setting each factor to zero gives the solutions x=13x = \frac{1}{3} and x=−2x = -2. The positive difference between these solutions is 13−(−2)=13+2=73\frac{1}{3} - (-2) = \frac{1}{3} + 2 = \frac{7}{3}.

Step-by-Step Solution

1
Rearrange the quadratic equation into standard form ax2+bx+c=0ax^2 + bx + c = 0.
3x2+5x−2=03x^2 + 5x - 2 = 0
Before factoring a quadratic equation, all terms must be moved to one side so that the equation is set to zero.
2
Factor the quadratic expression by finding two binomials that multiply to 3x2+5x−23x^2 + 5x - 2.
(3x−1)(x+2)=0(3x - 1)(x + 2) = 0
Since the product of 33 and −2-2 is −6-6, we look for factors of −6-6 that sum to 55, which are 66 and −1-1. Splitting the middle term yields 3x2+6x−x−2=3x(x+2)−1(x+2)=(3x−1)(x+2)3x^2 + 6x - x - 2 = 3x(x + 2) - 1(x + 2) = (3x - 1)(x + 2).
3
Use the zero-product property to solve for xx by setting each binomial factor to zero.
3x−1=0⇒x=133x - 1 = 0 \Rightarrow x = \frac{1}{3} and x+2=0⇒x=−2x + 2 = 0 \Rightarrow x = -2
If the product of two quantities is zero, at least one of the quantities must be zero.
4
Calculate the positive difference between the two solutions.
∣13−(−2)∣=∣13+2∣=73|\frac{1}{3} - (-2)| = |\frac{1}{3} + 2| = \frac{7}{3}
The positive difference is the absolute value of the subtraction of one root from the other.

Key Concept

Solving a quadratic equation by factoring over the integers after setting the equation equal to zero.

Alternative Method

Alternatively, the quadratic formula can be used. Once the equation is rewritten as 3x2+5x−2=03x^2 + 5x - 2 = 0, substitute a=3a=3, b=5b=5, and c=−2c=-2 into x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} to get x=−5±25−4(3)(−2)6=−5±76x = \frac{-5 \pm \sqrt{25 - 4(3)(-2)}}{6} = \frac{-5 \pm 7}{6}. This simplifies to x=13x = \frac{1}{3} and x=−2x = -2. The positive difference between these values is 13−(−2)=73\frac{1}{3} - (-2) = \frac{7}{3}.
Estimated Time:1m 30s
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Solving Quadratic Equations by Factoring

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