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

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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: $x = 14$ $x=14$

$\sqrt{x - 5} = \sqrt{2 x - 3} - 2$ $sqrt(x-5) = sqrt(2x-3)-2$

${(\sqrt{x - 5})}^{2} = {(\sqrt{2 x - 3} - 2)}^{2}$ $(sqrt(x-5))^2 = (sqrt(2x-3)-2)^2$

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

$- x - 6 = - 4 \sqrt{2 x - 3}$ $-x-6= -4sqrt(2x-3)$

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

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

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

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

$x = 6 or x = 14$ $x=6 or x=14$

Check for extraneous solutions:

Only x = 14 works in the original equation:

$\sqrt{14 - 5} = \sqrt{2 \cdot 14 - 3} - 2$ $sqrt(14-5) = sqrt(2*14-3)-2$

$3 = 3$ $3 = 3$

Hope that helped

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