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!

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