Unable to understand

[spacewater]

problem
$\displaystyle \sqrt {2x+6} - \sqrt{x+4} = 1$

I got this part .. to free $\displaystyle \sqrt{2x+6}$ by itself and square it, but I dont get how the right side of equation comes out to

$\displaystyle \sqrt{2x+6} = 1 + \sqrt{x+4}$

$\displaystyle 2x+6 = 1+ 2\sqrt{x+4} + (x+4)$ <------ I dont understand how this came out to be like this shouldnt it be $\displaystyle 2x + 6 = 1 + x + 4$ since we're squaring both side?

Can someone who isn't too busy elaborate this problem to me please?

 

[royhaas]

Start by simplifying so that the radical is on the right hand side.

 

[Denis]

$\displaystyle 2x+6 = 1+ 2\sqrt{x+4} + (x+4)$

Isolate square root term:
2SQRT(x + 4) = x + 1
Square again:
4(x + 4) = x^2 + 2x + 1

OK??

 

[Aladdin]

problem
$\displaystyle \sqrt {2x+6} - \sqrt{x+4} = 1$

I got this part .. to free $\displaystyle \sqrt{2x+6}$ by itself and square it, but I dont get how the right side of equation comes out to

$\displaystyle \sqrt{2x+6} = 1 + \sqrt{x+4}$

$\displaystyle 2x+6 = 1+ 2\sqrt{x+4} + (x+4)$ <------ I dont understand how this came out to be like this shouldnt it be $\displaystyle 2x + 6 = 1 + x + 4$ since we're squaring both side?------(a + b)^2 = a^2 + 2ab + b^2

Can someone who isn't too busy elaborate this problem to me please?