This one showed up in the middle of my quadratic equations. There is no y or 0. What am I supposed to do with it?
Give exact and approximate solutions to three decimal places.
(x+5)^2=81
x + 5 = + sqrt 81 = +9
or
x + 5 = - sqrt 81 = -9
so
x = 9-5 = 4
or
x = -9-5 = -14
check those back
How to check back:
if x = 4
what is (x+5)^2 ? hint 9^2 = 81 !!!
if x = -9
what is (x+5)^2 ? hint (-9)*(-9) = 81
we did it :)
perhaps it is coffee time
if x = -14
what is (x+5)^2 ? hint (-9)*(-9) = 81
we did it :)
Oh, and to do this as quadratic equation:
(x+5)^2 = 81
x^2 + 10 x + 25 = 81
x^2 + 10 x - 56 = 0
x = -10/2 +/- (1/2) sqrt(100 + 4*1*56)
x = -5 +/- (1/2)sqrt ( 324)
sqrt 324 = 18 by the way
x = -5 +/- 9
x = -5+9 = +4
or
x = -5-9 = -14