solve asap please

Question

Having trouble with 2 questions can you walk me through both? do you just square the radicals?

 

solve

 

√4x-2 = √2x+6

 

and...

 

√2x-1    +2  = √x+2    +3

 

Thanks!

Answer 1

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)