How Do You Solve This Quadratic Function By Completing The Square X^2+2x-7?

Question

Please Give Step By Step Instructions In Words On How You Solved It Please..Thanx..

o yea i forgot its actually X^2+2-7=0

Answer 1

X² + 2x - 7 = 0

add 1 to both sides

X² + 2x +1 - 7 = 1

(x + 1)² - 7 = 1

Add 7 to both sides

(x + 1)² = 8

take square root of both sides

x + 1 = ±2√2

subtract 1 from both sides

x = -1 ±2√2
.

Answer 2

The completion of the square follows from the form
b = (x+a)²
b = x² + 2ax + a² (expanded form)

So, you're looking for an a that allows you accomplish this form.
7 = x² + 2x
7 = x² + 2•1x

Using a = 1, complete the expanded form as
7 = x² + 2•1x + 1² - 1²
7 + 1 = x² + 2•1x + 1²
8 = (x + 1)²
√8 = √(x + 1)²
2√2 = |x + 1|
2√2 = x + 1 for x + 1 ≥ 0 or
2√2 = -(x + 1) for x + 1 < 0
2√2 - 1 = x for x ≥ ‾1 or
x = ‾2√2 - 1 for x < ‾1

Answer 3

x² +2x - 7 = 0

(x² + 2x +1 ) - 1- 7 = 0 ( took 2 from 2x, divided it by 2 , squared it, added and subtracted this number so it adds to 0)

(x +1)² - 8 = 0 (because (a +b)² = (a² +2ab + b²) )

(x + 1)²= 8 (because I am a genius ;))

x + 1 = ±√8 (took square root of both sides)

x = -1±2√2 (brought 1 to the other side)

so the two cases are:

x = -1 +2√2 and x = -1 - 2√2

Answer 4

x^2 + 2x + 1 = 7 + 1 = 8
(x+1)^2 = 8
x = -1 ± 2√2

Answer 5

if it is x^2 + 2 - 7 then it would be x^2 = 5 so x = +/- squareroot of 5. put that in your calc if you need a decimal answer.