Question

How do you solve `x^2+10x+16=0` by completing the squares?

Answer 1

`x = -8, -2`

Explanation:

Given: `x^2 + 10x + 16 = 0`

To complete the square put the two `x` terms on the left and the constant on the right of the equation:

`color(red)(x^2 + 10x) = -16`

Complete the square by multiplying the `x`-term by `1/2`:
`1/2 * 10 = 5`

and adding the square of this number to the right side of the equation: `5^2 = 25`

`(x +5)^2 = -16 color(blue)(+ 25)`

UNDERSTANDING CHECK:
`(x+5)^2 = color(red)(x^2 + 10x) color(blue)(+ 25).`

The `+25` was not in the original equation. If we add `+25` to one side of the equation, we must add the same amount to the other side of the equation to keep it balanced.

`(x+5)^2 = 9`

To solve, square root both sides of the equation:

`sqrt((x+5)^2) = +- sqrt(9)`

`x + 5 = +- 3`

`x = -5 +- 3`

`x = -5 + 3 = -2, " "x = -5 -3 = -8`

`x = -8, -2`

Answer 2

`x = -2 or -8`

Explanation:

The process of completing the square is done by adding a missing term to an expression so as to create the square of a binomial.

In `x^2 +10x+16=0," "16` is not the required constant, so move it to the right side.

`x^2 +10x" "=-16`

The required constant is determined from `color(blue)((b/2)^2)` where `b=10`

`color(blue)((10/2)^2 = 5^2 =25)`

Add this to both sides of the equation:

`x^2 +10xcolor(blue)(+25)=-16 color(blue)(+25)`

The left side is now the square of a binomial, ie a perfect square.

`" "(x+5)^2 = 9" "larr` find the square root of both sides.

`" "x+5 = +-sqrt9 = +-3`

`" "x = +-3-5`

This leads to two solutions:

`x = +3-5 = -2" "or" "x = -3-5=-8`