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Mathematics
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square both sides first... what do you get?
\[(\sqrt{6x-8})^2 = x^2\]
when you square a square root, you get back whatever is in the radicand...

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Other answers:

The radical is over the entire left side of the equation. If you square some number then take the square root of that same number its like you did nothing to that number. In this case think of the whole left side of the equation as some number.
\[ \left(\sqrt{\text{SOMETHING}}\right)^2 = \text{SOMETHING} \]
\(\large (\sqrt{radicand})^2 = radicand \)
\[x^2 = x^2 \]
\[6x-8 = x^2\]
What should you do next?
You could do that but you want to get the x's together because you are solving for x.
Well the most efficient thing to do would be to subtract 6x from both , right?
So whats next?
6x-6x = 0 Remember "like terms"
\[-8 = x^2 - 6x\]
If we have 6x that means six times x or we have six x's. We could say our x is equal to apples. If you have six apples one would need to take six apples away to get zero. Its the same concept here to get zero and therefore get ride of 6x from the left side we need to take six x's away. And so we don't change the meaning of the equation we have to take six x's away from both sides.
still need help?
ok.... you ended up with with \(\large -8=x^2-6x \)... this is a quadratic equation and to solve quadratic equations you must first get it in standard form: \(\large 0=ax^2+bx+c \) so all you need to do is move that -8 over to the right side: \(\large 0=x^2-6x+8 \)
bye... i'll continue...
one of the easiest way to solve a quadratic equation (when possible) is to factor: \(\large 0=x^2-6x+8 \) \(\large 0=(x-4)(x-2) \) from here its obvious that x=4 or x=2
but you must now check for extraneous roots against your original problem: \(\large \sqrt{6x-8}=x \) check both solutions to see which one works....

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