How do you solve #sqrt(3m+28) =m#?

1 Answer
Don't Memorise
Jul 20, 2015

#color(blue)( m=7#

Explanation:

#sqrt(3m+28) =m#

Squaring both sides
#(sqrt(3m+28))^2 =m^2#

#3m +28=m^2#

#m^2 - 3m -28=0#

Factorising by splitting the middle term (in order to find the solutions)**

#m^2 - color(green)(3m) -28=0#

#m^2 - color(green)(7m +4m) -28=0#

#m(m-7) +4 (m-7)=0#

#(m+4)(m-7) =0#

Upon equating the factors with zero we can obtain solutions:

#m+4=0, m=-4#
This solution is not applicable as square root of an expression cannot be negative

So the solution is
#(m-7) =0, m=7#

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