algebra, solve x

Quote:

Originally Posted by desmon

Find x.

$\frac{3x^2 + 5x - 2}{3x^2 + 2x} = 0$

Any help is greatly appreciated.

Thanks

$\frac{3x^2 + 5x - 2}{3x^2 + 2x} = 0$ inplies that $3x^2 + 5x - 2 = 0$ and that $3x^2 + 2x \not = 0$

Solving $3x^2 + 5x - 2 = 0$

$3x^2 + 5x - 2 = 0$
$3x^2 + 6x - x - 2 = 0$
$3x(x + 2) - (x + 2) = 0$
$(3x - 1)(x + 2) = 0$

So $x = \frac{1}{3}$ or $x = -2$

Checking the condition that $3x^2 + 2x \not = 0$

$3x^2 + 2x \not = 0$
$x(3x + 2) \not = 0$
$x \not = 0$ or $x \not = -\frac{2}{3}$