Question

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

Answer 1

x = -3,
`$ (3)/(2) $`

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 = (-b \pm √(b^2 - 4ac))/(2a) $`

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

Therefore, the roots of the equation are:

`$ x = (- 3 \pm √(9 - 4(2)(-9)))/(2(2)) $`

`$ \implies x = (-3 \pm √(81))/(4) $`

`$ \implies x = (-3 \pm 9)/(4) $`

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

x =
`$ (-3 + 9)/(4) $` and
`$ (- 3 - 9)/(4) $`

`$ \implies x = (6)/(4) \hspace{5mm} \& \hspace{5mm} (-12)/(4) $`

⇒ x = -3, 3/2

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

(x + 3)(x -
`$ (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 -
`$ (3)/(2) $` to see that this indeed is the given quadratic equation.