Unable to understand

spacewater

Junior Member
Joined
Jul 10, 2009
Messages
67
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?
 
Start by simplifying so that the radical is on the right hand side.
 
spacewater said:
\(\displaystyle 2x+6 = 1+ 2\sqrt{x+4} + (x+4)\)
Isolate square root term:
2SQRT(x + 4) = x + 1
Square again:
4(x + 4) = x^2 + 2x + 1

OK??
 
spacewater said:
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?
 
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