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

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

The sum of the distinct real solutions is

\frac{8}{3}

To solve

(2x - 1)^2 = (x + 2)^2

, we rearrange the equation to

(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

, which simplifies to

(x - 3)(3x + 1) = 0

. The solutions are

x = 3

and

x = -\frac{1}{3}

. Summing these gives

3 + (-\frac{1}{3}) = \frac{8}{3}

Rearrange the equation by moving all terms to one side to set it to zero.

(2x - 1)^2 - (x + 2)^2 = 0

To solve a quadratic equation by factoring, it must first be set equal to zero.

Factor the expression using the difference of squares formula,

a^2 - b^2 = (a - b)(a + b)

, where

a = 2x - 1

and

b = x + 2

[(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.

Simplify the terms inside each set of brackets.

(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.

Set each linear factor to zero to find the distinct real solutions.

x - 3 = 0 \Rightarrow x = 3

and

3x + 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.

Calculate the sum of the distinct real solutions.

3 + \left(-\frac{1}{3}\right) = \frac{9}{3} - \frac{1}{3} = \frac{8}{3}

The question asks for the sum of the solutions.

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

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