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How do I solve this? \[-7\sqrt[2]{3x-8} + 3 = -4x + 7\]
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\[-7\sqrt[2]{3x-8} + 3 = -4x + 7\]
\[-7\sqrt{3x-8}=-4x+4\\ \sqrt{3x-8}=\frac{-4x+7}{-7}\] then square both sides.
But there is a 2 infront of the square root of 2x-8 How do i get rid of that?
You want to start first by solving the squareroot.
Don't you think that 2 simply means square root.
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@imron07 is right. if there were a 3 there it would mean ^(1/3). The 2 just means ^(1/2) which is the same thing as square root
Oh that makes so much sense. Thanks guys, I really appreciate it.
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