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Find the non-extraneous solutions of the square root of x+6-5=x+1
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non-extraneous ?
yeah thats what the q says
here are the answer choices: x = −2 x = −6 x = −6 and x = −2 x = 2
is it \(\sqrt{ ( x+6)}-5=x+1\) ?
\(\sqrt{ ( x+6)} =x+6\) so x=-6, -4 not sure what is non-extraneous @ganeshie8 ?
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sorry i ment =-6,-2
yeah but without the peranthasis
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