Question

X^{2} - 8x + 7 = 0
Solve by completing the square. Also can you include steps?

Answer 1

`x = 7, 1`

Explanation:

`x^(2) - 8x + 7 = 0`

`x^(2) - 8x = -7`

`x^(2) - 8x + (-4)^(2) = -7+(-4)^(2)` → the
`-4^(2)` was obtained by doing
`(b)/(2) ^(2)`

`x^(2) - 8x + 16 = 9`

`(x - 4) = ± √(9)`

`(x - 4) = ± 3`

`x = 3 + 4` or
`x = -3 + 4`

Note:

What makes this a "complete-the-square" procedure? - Well, we created a perfect suqare when doing
`(b)/(2)^(2)` , and that is what makes our procedure a "complete-the-suqare" procedure.

Hope it helped,

BioTeacher101

Answer 2

{1, 7}

Explanation:

Start with x^2 - 8x + 7 = 0.

We want to modify the first two terms to resemble x^2 - ax + b - b, where a is the coefficient (here -8) of the x term and b is the square of half of a.

Here, a = -8. Half of that is -4. The square of -4 is 16, and this is the value of b.

Write out x^2 - 8x as x^2 - 8x + 16 - 16. This is the perfect square of -4, added to x^2 - 8x and then subtracted from the result

Then the original equation, x^2 - 8x + 7 = 0, can be rewritten as:

x^2 - 8x + 16 - 16 + 7 = 0, and then as (x - 4)^2 - 9, or (x - 4)^2 = 9.

We want to solve this result for x. To do this, take the square root of both sides:

x - 4 = ±√9, or x - 4 = ±3.

Adding 4 to both sides, we get:

x = 4 ± 3. Thus, the solutions are x = 4 + 3 = 7 and x = 4 - 3 = 1.