# Exponent problem

G

#### Guest

##### Guest
Hi guys, i am kindda new to maths and i need help urgently to understand some of exponential problem.
i have to solve this problem:
sqrt(1+2x+x^2) = 2
what i did:
1^(0.5)+ 2x^(0.5)+x= 2
2x^0.5+x = 2-1^0.5
then from here i am not sure to solve this problem do i have to do :roll:

#### daon

##### Senior Member
You can't raise individual terms to the one-half power in order to distribute a square root:

$$\displaystyle \L \sqrt{x^2+2x+1}=2 \Rightarrow (x^2+2x+1)^{\frac{1}{2}} = 2 \\$$

Now square both sides:

$$\displaystyle \L((x^2+2x+1)^{\frac{1}{2}})^{2} = 2^2\\ \Rightarrow (x^2+2x+1)^{\frac{2}{2}} = 4\\ \Rightarrow x^2+2x+1 = 4\\$$

Now use your knowledge of quadratics to solve.

#### Denis

##### Senior Member
if sqrt(a) = x
then a = x^2
(tatoo that under your right armpit!)

G

#### Guest

##### Guest
thanks for the explanation it help a lot ^-^

daon said:
You can't raise individual terms to the one-half power in order to distribute a square root:

$$\displaystyle \L \sqrt{x^2+2x+1}=2 \Rightarrow (x^2+2x+1)^{\frac{1}{2}} = 2 \\$$

Now square both sides:

$$\displaystyle \L((x^2+2x+1)^{\frac{1}{2}})^{2} = 2^2\\ \Rightarrow (x^2+2x+1)^{\frac{2}{2}} = 4\\ \Rightarrow x^2+2x+1 = 4\\$$

Now use your knowledge of quadratics to solve.