# How do you solve 9d^2-81=0?

Oct 2, 2017

You can factor out 9 immediately:

$9 \left({d}^{2} - 9\right)$ ...and then factor the rest, giving:

$9 \left(\left(d + 3\right) \left(d - 3\right)\right) = 0$, from which you can read off the values for d directly: $d = \pm 3$

Or, you can plug the coefficients from the original equation into the quadratic formula. Remember this will solve any quadratic having of form $a {x}^{2} + b x + c = 0$.

...substitute 'd's for x's, and use a = 9, b = 0, c = -81.

and plug 'em into the quadratic formula:

$d = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

remember that coefficient b is zero for this case, so we get:

$d = \pm \frac{\sqrt{- 4 \cdot 9 \cdot \left(- 81\right)}}{18}$

$d = + = \frac{\sqrt{2916}}{18} = \pm \frac{54}{18} = \pm 3$

(I think the first way is easier, but hey.)

GOOD LUCK!