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Mathematics 20 Online
OpenStudy (anonymous):

Please help me check if I am going the the right direction..solve squire-root of 2x+15=x+2

OpenStudy (anonymous):

no you should make the equation equals zero its more easier for you

OpenStudy (anonymous):

\[\sqrt{2x+15} = x+2\]

OpenStudy (anonymous):

okay check what i did

OpenStudy (anonymous):

try make it like this \[\sqrt{x + 13}\]

OpenStudy (anonymous):

=0

OpenStudy (anonymous):

\[\sqrt{2x+15}= x+2 =(\sqrt{2x+15)^{2}}\]

OpenStudy (anonymous):

what is that ?

OpenStudy (anonymous):

\[\sqrt{2x+15}=x+2 =2x+15=(x+2)^2, =2x+15=x^2+4x+4\]

OpenStudy (anonymous):

now i solve for x\[x^2+4x+4=2x+15=x^2+4x+4-2x-15=0\]

OpenStudy (anonymous):

=x^2+2x-11=0

OpenStudy (anonymous):

so the first WAS a root !!

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

that is another thing then but you wrote them both with no roots

OpenStudy (anonymous):

the question is\[\sqrt{2x+15}=x+2\]

OpenStudy (anonymous):

then square them to find x

OpenStudy (anonymous):

it`ll be x = {\[\frac{ -2+\sqrt{48} }{ 2 } , \frac{ -2-\sqrt{48} }{ 2 }\]}

OpenStudy (anonymous):

where did you get 48

OpenStudy (imstuck):

Actually, simplifying that you get \[-1\pm2\sqrt{3}\]

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