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

Question

1962 views around the world

You can reuse this answer
Creative Commons License

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)#