What is the positive difference between the two real solutions to the equation (2x - 1)^2 … — ACT

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

The positive difference between the two real solutions is 5.

The correct answer is 5. By expanding the equation, we get

4x^2 - 4x + 1 = 3x^2 - 5x + 7

. Rearranging the terms to set the equation to zero yields

x^2 + x - 6 = 0

. Factoring this equation gives

(x + 3)(x - 2) = 0

, which has the solutions

x = -3

and

x = 2

. The positive difference between these two solutions is

|2 - (-3)| = 5

Expand both sides of the equation.

4x^2 - 4x + 1 = 3x^2 - 5x + 7

Expanding the squared term on the left side and distributing the

x

on the right side allows us to write the equation in polynomial form.

Rearrange the equation by moving all terms to the left side to set the right side to zero.

x^2 + x - 6 = 0

Subtracting

3x^2

, adding

5x

, and subtracting

7

from both sides simplifies the equation into standard quadratic form,

ax^2 + bx + c = 0

Factor the quadratic trinomial.

(x + 3)(x - 2) = 0

Finding two integers that multiply to

-6

and add to

1

gives

3

and

-2

, allowing us to factor the equation over the integers.

Solve for

x

by setting each linear factor to zero.

x = -3

and

x = 2

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 positive difference between the two solutions.

|2 - (-3)| = 5

Subtracting the smaller solution from the larger solution gives the positive distance between them on the number line.

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

Estimated Time:2m 0s

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