Solve for X: 2 = √((1+x)² + (1-2x)²)?

Question

How do you go about solving this? I'm stumped as to how to get anything out of the radical.

The solution given is (1±√11)/5

Answer 1

FOIL 2=√(1+2x+x^2 + 1 – 4x + 4x^2)
Like Terms 2=√(2-2x+5x^2)
Square Both Sides 4=2-2x+5x^2
Set equal to zero in descending order 0=5x^2-2x-2
Factor using quadratic formula x= {-(b) ±√[(b) ^2-4(a)(c)]}/2(a)
x= {-(-2) ±√[(-2) ^2-4(5)(-2)]}/2(5)
= [2±√(4+40)]/10
= (2±√44)/10
= {2±√[4(11)]}/10
= (2±2√11)/10
=(1±√11)/5

Answer 2

Square both sides, then you get a quadratic, which you can use the quadratic formula on.

More specifically, squaring both sides gives you:
4 = (1+x)^2 + (1-2x)^2
4 = 1 + 2x + x^2 + 1 - 4x + 4x^2
5x^2 - 2x - 2 = 0

Then the quadratic formula gives you:
[2 +/- sqrt(44)]/10 =
[1 +/- sqrt(11)]/5

Answer 3

2 = √((1+x)^2 + (1-2x)^2)
2 = √(x^2 + 1 + 2x + 4x^2 + 1 - 4x)
2 = √(x^2 + 1 + 4x^2 + 1 - 2x)
2 = √(5x^2 - 2x + 2)
5x^2 - 2x + 2 = 4 (Square both sides)
5x^2 - 2x - 2 = 0
Apply quadratic formula:

x = [2 ±(√4 + 40)]/10
x = (2 ±√44)/10
x = (2 ±2√11)/10
x = 2(1 ±√11)/ 2*5
x = (1 ±√11)/5

Answer 4

4 = (1 + x) ² + (1 - 2x) ²
4 = 1 + 2x + x² + 1 - 4x + 4x²
5x² - 2x - 2 = 0
x = [2 ± √(4 + 40)] / 10
x = (2 ± 2√11) / 10
x = (1/5) (1 ±√11)

Answer 5

2 = √((1+x)² + (1-2x)²)
4 =(1+x)² + (1-2x)²
4 = 1 + x² + 2x + 1 + 4x² - 4x
5x² -2x - 2 = 0
a = 5 , b = -2 c = -2

x = [-(-2) ± √((-2)²-4(5)(-2))] / 2(5)
x = [ 2 ± √(4 + 40)] / 2(5)
x = [ 2 ± √44] / 2(5)
x =(1±√11)/5

Answer 6

square both sides.
that's it.