√8x+1 = 5
√5x-4 = 6
√7x-3 -2 = 0 , the -2 is not under the radical sign
√5x^2 - 3x = 2x
√8x+1 = 5
√5x-4 = 6
√7x-3 -2 = 0 , the -2 is not under the radical sign
√5x^2 - 3x = 2x
√(8x + 1) = 5 ---> (√(8x + 1)^{2} = 5^{2} ---> 8x + 1 = 25 ---> 8x = 24 --->
x = 2
√(5x - 4) = 6 ---> (√5x - 4)^{2} = 6^{2} ---> 5x -4 = 36 ---> 5x = 40 --->
x = 8
√(7x - 3) - 2 = 0 ---> (√(7x - 3))^{2} = 2^{2} ---> 7x - 3 = 4 --->
x = 1
√(5x^{2}) - 3x = 2x ---> √(5x^{2}) = 5x (x ≥ 0) ---> x · √5 = 5x ---> √5 ≠ 5 - given equation does not have any solutions.
√(5x^{2} - 3x) = 2x (x ≥ 0) ---> (√(5x^{2} -3x))^{2} = (2x)^{2} ---> 5x^{2} -3x = 4x^{2} --->
(5x^{2} - 4x^{2}) -3x = 0 ---> x^{2} -3x = 0 ---> x(x - 3) = 0 --->
x_{1} = 0 , x_{2} = 3
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
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.
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