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sqrt[8]{x-3}=sqrt[4]{x+2}?
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square both sides and temme wut u get afta that
(x-3)^8=(x+2)^4??
\[{8(x-3)}^1/2={4(x+2)}^1/2\]
is it this way
\[\sqrt{8}(x-3)=\sqrt{4}(x+2)\] is this the question?
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well if its what i wrote then first u distribute the sqrt 8 and sqrt4 \[\sqrt8 x-3\sqrt8 = \sqrt4 x +2\sqrt4\] but\[\sqrt8=\sqrt4 \times \sqrt2 =2\sqrt 2\]and\[\sqrt4 =2\] so the equation becomes \[2\sqrt2 x-(3 \times 2\sqrt2) = 2x+ (2 \times 2)\]\[=2\sqrt2 x -6\sqrt2 = 2x+4\]\[2\sqrt2 x-2x=4+6\sqrt2\]\[x(2\sqrt2 -2) = 4+6\sqrt 2\]\[x=\frac{4+6\sqrt2}{2\sqrt2 -2}\]
u just need to simplify now
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