Question

2x^2=9-3x Find the factors

Answer 1

Answer:

x = -3, $\frac{3}{2}$

Step-by-step explanation:

The given quadratic equation is: $2x^2 = 9 - 3x$

This can be written as: $2x^2 + 3x - 9 = 0$

To solve a quadratic equation of the form $ax^2 + bx + c = 0$ we use the formula:

           $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, a = 2; b = 3; c = - 9

Therefore, the roots of the equation are:

$x = \frac{- 3 \pm \sqrt{9 - 4(2)(-9)}}{2(2)}$

$\implies x = \frac{-3 \pm \sqrt{81}}{4}$

$\implies x = \frac{-3 \pm 9}{4}$

We get two values of 'x', viz.,

x = $\frac{-3 + 9}{4}$ and $\frac{- 3 - 9}{4}$

$\implies x = \frac{6}{4} \hspace{5mm} \& \hspace{5mm} \frac{-12}{4}$

⇒ x = -3, 3/2

Since the factors of the quadratic equation is asked, we write it as:

(x + 3)(x - $\frac{3}{2}$) = 0

because, if (x - a)(x - b) are the factors of a quadratic equation, then 'a' and 'b' are its roots.

Multiply (x + 3) and (x - $\frac{3}{2}$ to see that this indeed is the given quadratic equation.