How do you solve #x( 5x - 2) = 24#?

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Answer 1 · 2018-05-01T21:18:19

#x=12/5 or 2.4 , -2#

expand:

#x(5x-2) -> 5x^2-2x=24#

set equal to #0#:

#5x^2-2x=24 -> 5x^2-2x-24=0#

factorise:

#-> 5x^2+10x-12x-24#

#-> 5x(x+2)-12(x+2)#

#rArr (x+2)(5x-12)#

Check to see if correct:

#(5x-12)(x+2)=5x^2+10x-12x-24#

#-> 5x^2-2x-24#.

set each bracket equal to #0#:

#5x-12=0#

#-> 5x=12#

#rArr x=12/5#

#x+2=0#

#rArr x=-2#

Check these values:

#2.4(5 xx 2.4-2)=2.4 xx 10=24#

#-2(5 xx-2-2)#=-2 xx -12=24#

#therefore# these values are correct.