If 3x(x - 3) = 2(x + 3) - 12, what is the product of the two solutions to this equation? — ACT

If $3x(x - 3) = 2(x + 3) - 12$ , what is the product of the two solutions to this equation?

Cevap: 2

The product of the two solutions to the equation is 2.

The correct answer is 2. Expanding both sides of the equation

3x(x - 3) = 2(x + 3) - 12

gives

3x^2 - 9x = 2x + 6 - 12

, which simplifies to

3x^2 - 9x = 2x - 6

. Subtracting

2x - 6

from both sides results in the standard form quadratic equation

3x^2 - 11x + 6 = 0

. Factoring this expression gives

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

. Setting each factor to zero yields the solutions

x = \frac{2}{3}

and

x = 3

. Multiplying these solutions gives

\frac{2}{3} \times 3 = 2

. Alternatively, by Vieta's formulas, the product of the roots of a quadratic equation

ax^2 + bx + c = 0

\frac{c}{a}

, which directly gives

\frac{6}{3} = 2

Expand the expressions on both sides of the equation.

3x^2 - 9x = 2x + 6 - 12

, which simplifies to

3x^2 - 9x = 2x - 6

To prepare the equation for rearrangement into the standard quadratic form.

Move all terms to the left side of the equation to set it equal to zero.

3x^2 - 11x + 6 = 0

To write the quadratic equation in the standard form

ax^2 + bx + c = 0

Factor the quadratic trinomial by grouping.

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

To break down the quadratic equation into linear factors that can be solved individually.

Set each linear factor to zero and solve for

x

x = \frac{2}{3}

and

x = 3

According to the zero product property, if a product of factors is zero, at least one factor must be zero.

Calculate the product of the two solutions.

\frac{2}{3} \times 3 = 2

To find the product of the solutions as requested by the question.

Solving quadratic equations by factoring after expanding and rearranging terms

Soru