√8x+1 = 5

√5x-4 = 6

√7x-3 -2 = 0 , the -2 is not under the radical sign

√5x^2 - 3x = 2x

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√8x+1 = 5

√5x-4 = 6

√7x-3 -2 = 0 , the -2 is not under the radical sign

√5x^2 - 3x = 2x

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√(8x + 1) = 5 ---> (√(8x + 1)^{2} = 5^{2} ---> 8x + 1 = 25 ---> 8x = 24 --->
**x = 2
**√(5x - 4) = 6 ---> (√5x - 4)

√(7x - 3) - 2 = 0 ---> (√(7x - 3))

√(5x

√(5x

(5x

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

√(8x+1) = 5

Square both sides,

8x+1 = 25

x = 3 <==Answer

I don't understand how you got x=3

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## Comments

In the first problem, what happened to the 1?

8x+1 was under the radical sign.

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.

Thank you, so much.

You're very welcome, Kristine, but I made mistake, in first equation

x=3.Comment