How do you solve $\sqrt{3 x - 3} + \sqrt{2 x + 8} + 1 = 0$ $sqrt( 3x-3) + sqrt(2x+8) +1=0$ ?

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Answer 1 · 2016-04-06T00:30:00

Isolate one of the radicals on one side of the equation.

$\sqrt{3 x - 3} = - 1 - \sqrt{2 x + 8}$ $sqrt(3x - 3) = -1 - sqrt(2x + 8)$

${(\sqrt{3 x - 3})}^{2} = {(- 1 - \sqrt{2 x + 8})}^{2}$ $(sqrt(3x - 3))^2 = (-1 - sqrt(2x + 8))^2$

$3 x - 3 = 1 + 2 x + 8 + 2 \sqrt{2 x + 8}$ $3x - 3 = 1 + 2x + 8 + 2sqrt(2x + 8)$

$x - 12 = 2 \sqrt{2 x + 8}$ $x - 12 = 2sqrt(2x + 8)$

${(x - 12)}^{2} = {(2 \sqrt{2 x + 8})}^{2}$ $(x - 12)^2 = (2sqrt(2x + 8))^2$

$x^{2} - 24 x + 144 = 4 (2 x + 8)$ $x^2 - 24x + 144 = 4(2x + 8)$

$x^{2} - 24 x + 144 - 8 x - 32 = 0$ $x^2 - 24x + 144 - 8x - 32 = 0$

$x^{2} - 32 x + 112 = 0$ $x^2 - 32x + 112 = 0$

$(x - 28) (x - 4) = 0$ $(x - 28)(x - 4) = 0$

$x = 28 and 4$ $x = 28 and 4$

Checking both these solutions in the original equation we find that neither work. Therefore, there is no solution $(\emptyset)$ $(O/)$

Hopefully this helps!

How do you solve √3x−3+√2x+8+1=0sqrt( 3x-3) + sqrt(2x+8) +1=0?