Question

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

Answer 1

Like so:

Explanation:

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!