How do you solve #sqrt(3x+4)-sqrt(2x-7)=3#?

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Answer 1 · 2017-05-30T13:25:45

#x in {4, 64}#

We have:

#sqrt(3x +4) = 3 + sqrt(2x- 7)#

Squaring both sides, we get:

#(sqrt(3x+ 4))^2 = (3 + sqrt(2x - 7))^2#

#3x + 4 = 9 + 6sqrt(2x- 7) + 2x - 7#

Regrouping non-square root terms to one side of the equation, we get:

#x + 2 = 6sqrt(2x- 7)#

Square again:

#x^2 + 4x + 4 = 36(2x- 7)#

#x^2 +4x + 4 = 72x - 252#

#x^2 - 68x + 256 =0#

#(x -4)(x -64) = 0#

#x= 4 or 64#

Now test to see whether or not the solutions are valid.

Testing #x = 4#

#sqrt(3(4) + 4) =^? 3 + sqrt(2(4) - 7)#

#4 = 3 + 1 color(green)(√)#

Testing #x = 64#

#sqrt(3(64) + 4) =^? 3 + sqrt(2(64) - 7)#

#sqrt(196) =^? 3 + sqrt(121)#

#14 = 3 + 11 color(green)(√)#

So our solution set is #x in {4, 64}#.

Hopefully this helps!