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√(8x + 1) = 5 ---> (√(8x + 1)2 = 52 ---> 8x + 1 = 25 ---> 8x = 24 ---> x = 2
√(5x - 4) = 6 ---> (√5x - 4)2 = 62 ---> 5x -4 = 36 ---> 5x = 40 ---> x = 8
√(7x - 3) - 2 = 0 ---> (√(7x - 3))2 = 22 ---> 7x - 3 = 4 ---> x = 1
√(5x2) - 3x = 2x ---> √(5x2) = 5x (x ≥ 0) ---> x · √5 = 5x ---> √5 ≠ 5 - given equation does not have any solutions.
√(5x2 - 3x) = 2x (x ≥ 0) ---> (√(5x2 -3x))2 = (2x)2 ---> 5x2 -3x = 4x2 --->
(5x2 - 4x2) -3x = 0 ---> x2 -3x = 0 ---> x(x - 3) = 0 ---> x1 = 0 , x2 = 3

In the first problem, what happened to the 1?

3/16/2013

8x+1 was under the radical sign.

3/16/2013

Sorry Kristine for misreading the problem you posted. You have to use parentheses v(8x + 1). Even if there is square root from the product you still have to put parentheses, because there is no line above which will indicate the under radical expression.
About 1: it was subtracted from both sides of equation.

3/17/2013

Thank you, so much.

3/17/2013

You're very welcome, Kristine, but I made mistake, in first equation x=3.

3/17/2013

I show you how to do the first. You can try the following problems.

√(8x+1) = 5

Square both sides,

8x+1 = 25