How do you solve $sqrt(6x^2 + 5x) = 2$ ?

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Answer 1 · 2018-07-29T12:12:40

$x=1/2$ or $x=-4/3$

$sqrt(6x^2+5x)=2$

Square both sides
$6x^2+5x=4$

$6x^2+5x-4=0$

Using the quadratic formula,
$x=(-5+-sqrt(25+96))/12$

$x=(-5+-sqrt121)/12$

$x=(-5+-11)/12$

$x=(-5+11)/12$ or $x=(-5-11)/12$

$x=1/2$ or $x=-4/3$

$6x^2+5x-4=(x-1/2)(x+4/3)=(2x-1)(3x+4)$

How do you solve sqrt(6x^2 + 5x) = 2sqrt(6x^2 + 5x) = 2?