Question

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

Answer 1

You can factor out 9 immediately:

`9(d^2-9)` ...and then factor the rest, giving:

`9((d+3)(d-3)) = 0`, from which you can read off the values for d directly: `d = +-3`

Or, you can plug the coefficients from the original equation into the quadratic formula. Remember this will solve any quadratic having of form `ax^2 + bx + c = 0`.

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

and plug 'em into the quadratic formula:

`d = (-b +- sqrt(b^2-4ac))/(2a)`

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

`d = +-sqrt(-4*9*(-81))/18`

`d = +=sqrt(2916)/18 = +-54/18 = +-3`

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

GOOD LUCK!