SOLUTION: sqrt(4x-3) = 2+sqrt(2x-5) how to solve equations with two radicals

Question 1088968: sqrt(4x-3) = 2+sqrt(2x-5)
how to solve equations with two radicals

Found 3 solutions by MathLover1, ikleyn, natolino_2017:
Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!

sqrt(4x-3 2+sqrt(2x-5=> this is not good way to post a problem
I guess, you have something like this:
...if yes, than square both sides

....square both sides

....factor

solutions:
or

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
There is NO equation in you post.

There is incorrectly written expression only in your post.

Answer by natolino_2017(77)   (Show Source): You can put this solution on YOUR website!
First of all, we need to set the restriction to allow the expression to be well defined.
4x - 3 >=0 and 2x - 5 >=0
x >= 3/4 and x>=5/2
Restriction: x>=5/2
now we solve by squaring both sides:
4x-3 = 4 +4sqrt(2x - 5) + (2x-5)
2x-2 = 4sqrt(2x-5) = 2(x-1)
(x-1) is positive according to the restriction, so we can squaring both sides
4(x^2-2x+1) = 16(2x-5)
x^2-2x+1 = 8x-20
x^2-10x+21 = 0
(x-3)(x-7) = 0
so acoording to the last expression we have 2 solution x={3,7}
but according to the restriction x>=5/2 so we don't have to discard any solution.
Final Solution={3,7}

@natolino_

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