How do you solve $\sqrt{35 x} + 120 = x + 12$ $sqrt(35x) + 120 = x+12$ ?

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How do you solve $\sqrt{35 x} + 120 = x + 12$ $sqrt(35x) + 120 = x+12$ ?

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Answer 1 · 2015-04-23T00:07:29

Move all terms except the square root to one side of the equation and leave the square root on the other.
Square both sides.
Simplify and (hopefully) solve the resultant quadratic.

$\sqrt{35 x} + 120 = x + 12$ $sqrt(35x) + 120 = x+12$

$\sqrt{35 x} = x - 108$ $sqrt(35x) = x-108$

$35 x = x^{2} - 216 x + 108^{2}$ $35x = x^2 -216x +108^2$

$x^{2} - 251 x + 11664 = 0$ $x^2 - 251x + 11664= 0$

If I haven't made a mistake, things get a bit ugly...

using the formula for quadratic roots
$\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b+-sqrt(b^2-4ac))/(2a)$

we get the factors
$(x - 61.576) (x - 189.424)$ $(x-61.576)(x-189.424)$

That is
$x = - 61.576$ $x= -61.576$
or
$x = - 189.424$ $x= -189.424$

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How do you solve $\sqrt{35 x} + 120 = x + 12$ $sqrt(35x) + 120 = x+12$ ?

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How do you solve $\sqrt{35 x} + 120 = x + 12$ $sqrt(35x) + 120 = x+12$ ?