Answer:
x = -3, [tex]$ \frac{3}{2} $[/tex]
Step-by-step explanation:
The given quadratic equation is: [tex]$ 2x^2 = 9 - 3x $[/tex]
This can be written as: [tex]$ 2x^2 + 3x - 9 = 0 $[/tex]
To solve a quadratic equation of the form [tex]$ ax^2 + bx + c = 0 $[/tex] we use the formula:
[tex]$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $[/tex]
Here, a = 2; b = 3; c = - 9
Therefore, the roots of the equation are:
[tex]$ x = \frac{- 3 \pm \sqrt{9 - 4(2)(-9)}}{2(2)} $[/tex]
[tex]$ \implies x = \frac{-3 \pm \sqrt{81}}{4} $[/tex]
[tex]$ \implies x = \frac{-3 \pm 9}{4} $[/tex]
We get two values of 'x', viz.,
x = [tex]$ \frac{-3 + 9}{4} $[/tex] and [tex]$ \frac{- 3 - 9}{4} $[/tex]
[tex]$ \implies x = \frac{6}{4} \hspace{5mm} \& \hspace{5mm} \frac{-12}{4} $[/tex]
⇒ x = -3, 3/2
Since the factors of the quadratic equation is asked, we write it as:
(x + 3)(x - [tex]$ \frac{3}{2} $[/tex]) = 0
because, if (x - a)(x - b) are the factors of a quadratic equation, then 'a' and 'b' are its roots.
Multiply (x + 3) and (x - [tex]$ \frac{3}{2} $[/tex] to see that this indeed is the given quadratic equation.
