How do you solve the equation #9x^2+30x+25=11# by completing the square?

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Answer 1 · 2017-09-14T12:57:22

Like so:

For starters, the #x^2# term needs to be a multiple of 1, not 9.
So the first step is to divide both sides by 9, giving:

#x^2 + 30/9x + 25/9 = 11/9#

For me, it makes some of the subsequent steps easier if we subract #25/9# from both sides...

#x^2 + 30/9x = 11/9 - 25/9 = -14/9# <- we'll call this equation 1

Now remember how binomials are squared:

#(x + a)^2 = x^2 + 2ax + a^2#

So, if #30/9 = 2a#, then #a = 1/2 * 30/9 = 15/9#

...and #a^2 = 225/81#

so, what we do is, add and subtract this to the left side of eq. 1:

#x^2 + 30/9x + 225/81 - 225/81 = -14/9#

...and now the first 3 terms of the left side are a perfect square:

#(x + 15/9)^2 - 225/81 = -14/9#

...add #225/81# to both sides:

#(x + 15/9)^2 = -14/9 + 225/81 = (-126 + 225)/81 = 99/81#

...square root of both sides:

#x + 15/9 = sqrt(99/81)#

subtract #15/9# from both sides:

#x = sqrt(99/81) - 15/9#

#x = (sqrt(99) - 15)/9#

GOOD LUCK!