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

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


   

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) About Me  (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:
sqrt%284x-3%29=+2%2Bsqrt%282x-5%29...if yes, than square both sides
%28sqrt%284x-3%29%29%5E2=+%282%2Bsqrt%282x-5%29%29%5E2
4x-3=+4%2B4sqrt%282x-5%29%2B%28sqrt%282x-5%29%29%5E2
4x-3=+4%2B4sqrt%282x-5%29%2B2x-5
4x-2x%2B5-4-3=+4sqrt%282x-5%29
2x-2=+4sqrt%282x-5%29
cross%282%29%28x-1%29=+cross%284%292sqrt%282x-5%29
%28x-1%29=+2sqrt%282x-5%29....square both sides
%28x-1%29%5E2=+%282sqrt%282x-5%29%29%5E2
x%5E2-2x%2B1+=+4%282x-5%29
x%5E2-2x%2B1+=+8x-20
x%5E2-2x%2B1-8x%2B20=0
x%5E2-10x%2B21=0....factor
x%5E2-7x-3x%2B21=0
%28x%5E2-7x%29-%283x-21%29=0
x%28x-7%29-3%28x-7%29=0

%28x-3%29%28x-7%29=0
solutions:
x=3 or x=7


Answer by ikleyn(52781) About Me  (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) About Me  (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_