Casey W. answered • 06/18/15

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Mathematics (and Science) Instruction by a Mathematician!

You cannot simply square radicals in any equation any way you want...when you have an equality, a legal operation is to square both sides of the equation.

In the first problem it is sometimes hard to discern what it says when we use the √ symbol, I often use \sqrt{4x} as we would in LaTeX, or simply sqrt(4x) or best yet (4x)^(1/2) or (4x)^.5...this last interpretation is algebraically best...it indicates that the square root function is precisely the function that raises a number to the exponent one-half.

Re-writing your first problem in this notation:

(4x)^.5 - 2 = (2x)^.5+6

We next want to separate on different sides of the equation, all terms with x from all constant terms, as follows:

(4x)^.5 - (2x)^.5 = 6+2

Notice that there is a common factor of x^.5 on the LHS, lets factor this out:

(x)^.5(4^.5-2^.5) = 8

We can recognize 4^.5=2 (since the square root of 4 is 2), and then divide by (2-2^.5) to write:

x^.5 = 8/(2-2^.5)

NOW WE CAN SQUARE BOTH SIDES OF THE EQUATION:

(x^.5)^2 = [8/(2-2^.5)]^2

And since square root of x squared is x, we have solved for x explicitly. You can clean this up by squaring the RHS, and write:

x= 64/(6-4√2)

Josh F.

Yup it was your 2nd comment that was correct thanks

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06/18/15

Casey W.

06/18/15