What is the positive difference between the two real solutions to the quadratic equation 2… — ACT

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

The positive difference between the two solutions is

\frac{5}{6}

The correct positive difference is

\frac{5}{6}

because rearranging the equation

2x(3x - 5) = 3(x - 2)

yields

6x^2 - 13x + 6 = 0

. Factoring this equation gives

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

, which has the solutions

x = \frac{3}{2}

and

x = \frac{2}{3}

. Subtracting these values gives

\frac{3}{2} - \frac{2}{3} = \frac{9 - 4}{6} = \frac{5}{6}

Expand both sides of the equation.

6x^2 - 10x = 3x - 6

To begin solving, distribute the term

2x

on the left side and the constant

3

on the right side.

Rearrange the terms to set the quadratic equation to zero.

6x^2 - 13x + 6 = 0

Subtract

3x

and add

6

to both sides of the equation to write it in standard form

ax^2 + bx + c = 0

Factor the quadratic equation over the integers.

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

Find two binomials whose product is

6x^2 - 13x + 6

. We search for two numbers that multiply to

36

(from

6 \times 6

) and sum to

-13

, which are

-9

and

-4

, allowing factoring by grouping.

Solve for the roots of the equation.

x = \frac{3}{2}

x = \frac{2}{3}

Set each linear factor equal to zero using the Zero Product Property:

2x - 3 = 0 \implies x = \frac{3}{2}

and

3x - 2 = 0 \implies x = \frac{2}{3}

Calculate the positive difference between the two solutions.

\frac{3}{2} - \frac{2}{3} = \frac{9}{6} - \frac{4}{6} = \frac{5}{6}

Subtract the smaller solution from the larger solution to find the positive difference.

Solving quadratic equations by rearranging and factoring over integers.

Question