How do you solve the equation: $x=sqrt(3x+40)$ ?

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Answer 1 · 2018-05-28T13:19:31

$x=8$

$x = sqrt{3x + 40}$

Over reals the radical sign refers to the principal square root, so squaring both sides may introduce extraneous solutions.

$x^2 = 3x+40$

$x^2 - 3x -40 = 0$

$(x-8)(x+5)=0$

$x=8 or x=-5$

The principal square root won't ever be $-5$ so we reject that as extraneous.

$x=8$

Check:

$sqrt{3(8)+40}=sqrt{64}=8 quad sqrt$

How do you solve the equation: x=sqrt(3x+40)x=sqrt(3x+40)?