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

1 Answer
Oct 2, 2017

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!