How do you solve $sqrt(x-5) - sqrt(2x-3) = -2$ ?

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Answer 1 · 2015-10-14T00:52:51

Move one of the radicals to the other side of the equation, then square both sides a couple of times. Answer: $x=14$

$sqrt(x-5) = sqrt(2x-3)-2$

$(sqrt(x-5))^2 = (sqrt(2x-3)-2)^2$

$x-5 = 2x+1-4sqrt(2x-3)$

$-x-6= -4sqrt(2x-3)$

$(-x-6)^2= (-4sqrt(2x-3))^2$

$x^2+12x+36 = 32x-48$

$x^2-20x+84=0$

$(x-6)(x-14)=0$

$x=6 or x=14$

Check for extraneous solutions:

Only x = 14 works in the original equation:

$sqrt(14-5) = sqrt(2*14-3)-2$

$3 = 3$

Hope that helped

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