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Please help me check if I am going the the right direction..solve squire-root of 2x+15=x+2
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no you should make the equation equals zero its more easier for you
\[\sqrt{2x+15} = x+2\]
okay check what i did
try make it like this \[\sqrt{x + 13}\]
=0
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\[\sqrt{2x+15}= x+2 =(\sqrt{2x+15)^{2}}\]
what is that ?
\[\sqrt{2x+15}=x+2 =2x+15=(x+2)^2, =2x+15=x^2+4x+4\]
now i solve for x\[x^2+4x+4=2x+15=x^2+4x+4-2x-15=0\]
=x^2+2x-11=0
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so the first WAS a root !!
yes
that is another thing then but you wrote them both with no roots
the question is\[\sqrt{2x+15}=x+2\]
then square them to find x
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it`ll be x = {\[\frac{ -2+\sqrt{48} }{ 2 } , \frac{ -2-\sqrt{48} }{ 2 }\]}
where did you get 48
Actually, simplifying that you get \[-1\pm2\sqrt{3}\]
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