Question

What is x^2-8x-20=0 solving by completing the square?

Answer 1

`x=10`

Explanation:

`x^2-8x-20 = 0`

Add 20 to both sides...

`x^2-8x = 20`

When completed we should have a function of the form `(x+a)^2`. This function expanded would be `x^2+2ax+a^2`. If `2ax=-8x`, then `a=-4`, meaning our term will be `(x-4)^2`. Expanded this would give us `x^2-8x+16`, so to complete the square we have to add 16 to both sides...

`x^2-8x+16 = 20+16`

Now change it into our `(x+a)^2` form...

`(x-4)^2 = 36`

Square root both sides:

`x-4 = 6`

And finally add 4 to both sides to isolate x.

`x=10`

Answer 2

`x=10,\qquad\qquad x=-2`

Explanation:

First, move the `c` value to the RHS:

`x^2-8x=20`

Add `(\frac{b}{2})^2` to both sides:

`x^2-8x+(\frac{-8}{2})^2=20+(\frac{-8}{2})^2`

Simplifying the fractions:

`x^2-8x+16=20+16`

Now that the LHS is a perfect square, we can factor it as `(x-\frac{b}{2})^2`

`(x-4)^2=36`

Taking the real (non-principal) square root:

`\sqrt{(x-4)^2}=\sqrt{36}`

Simplifying:

`x-4=\pm 6`

Isolating for `x`:

`x=\pm 6+4`

`\quad x=-6+4,\qquad x=6+4`

`\therefore x=-2,\qquad\qquad x=10`