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:

Thanks in advance
 

daon

Senior Member
Joined
Jan 27, 2006
Messages
1,284
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
Joined
Feb 17, 2004
Messages
1,495
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.
 
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