How do you solve #sqrt(4x-3)= 2+sqrt( 2x-5)#?

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Answer 1 · 2016-01-31T06:58:56

#x = 7# and #x = 3#

When dealing with radicals on both sides of the equation, the first step is squaring both sides:

#(sqrt(4x-3))^2 = (2+sqrt(2x-5))^2#

Now we simplify:

#4x-3 = 4+4sqrt(2x-5)+2x-5#

#2x-2 = 4sqrt(2x-5)#

We square both sides again to get rid of the final radical:

#(2x-2)^2 = 16(2x-5)#

More simplifying:

#4x^2-8x+4 = 32x-80#

#4x^2-40x+84 = 0#

Solving the quadratic:

#4(x^2-10x+21) = 0#

#x^2-10x+21 = 0#

#(x-7)(x-3) = 0#

#x = 7# and #x = 3#

When substituting these answers into the original equation, you will find they check out.