Question

How do you solve `x^2 + 10x + 24 = 0` using completing the square?

Answer 1

`0 = x^2+10x+24 = (x+5)^2 - 1`

Hence `x = -5+-sqrt(1) = -5+-1`

Explanation:

`(x+5)^2 = x^2+10x+25`

So we have:

`0 = x^2+10x+24 = (x+5)^2 - 1`

Add `1` to both ends to get:

`(x+5)^2 = 1`

So:

`x+5 = +-sqrt(1) = +-1`

Subtract `5` from both sides to get:

`x = -5+-1`

That is `x=-6` or `x=-4`

Answer 2

Factor `y = x^2 + 10x + 24` by completing the square

Explanation:

`y = x^2 10x + (25 - 25) + 24 = 0`
`y = (x + 5)^2 - 1 = 0`
`(x + 5)^2 = 1` -> `x + 5 = +- 1`

x = -5 + 1 = -4
x = -5 - 1 = -6

Answer 3

Create a perfect square trinomial on the left side of the equation, then factor it and solve for `x`. The general equation for a perfect square trinomial is `a^2+2ab+b^2=(a+b)^2`.

Explanation:

`x^2+10x+24=0`

We are going to create a perfect square trinomial on the left side of the equation, then solve for `x`. The general equation for a perfect square trinomial is `a^2+2ab+b^2=(a+b)^2`.

Subtract `24` from both sides.

`x^2+10x=-24`

Divide the coefficient of the `x` term by `2`, then square the result, and add it to both sides.

`10/2=5` `; 5^2=25`

`x^2+10x+25=-24+25` =

`x^2+10x+25=1`

We now have a perfect square trinomial on the left side, in which `a=x and b=5`. Factor the trinomial, then solve for `x`.

`(x+5)^2=1`

Take the square root of both sides.

`x+5=+-sqrt1` =

`x=+-sqrt1-5` =

`x=sqrt 1-5=1-5=-4`

`x=-sqrt1-5=-1-5= -6`

`x=-4`
`x=-6`