How do you solve #x( 3x - 29) = - 26#?

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Answer 1 · 2018-04-04T15:21:50

#x=26/6, x=1#

Expand out.
#x(3x-29)=-26#
#3x^2-29x=-26#

Move -26 over, gives a quadratic equation
#3x^2-29x+26=0#

Factorise / graphically derive roots
#(3x-26)(x-1)#
graph{3x^2-29x+26 [-10, 10, -5, 5]}

#x=26/3, x=1#

Answer 2 · 2018-04-04T15:23:53

#x = 26/3 or x = 1 #

#3x^2 - 29x = -26#

#3x^2 - 29x - (-26) = -26 (-26)#

#3x^2 - 29x + 26 = 0#

#(3x - 26)(x-1) = 0#

So

# 3x−26=0 or x−1=0#

#x= 26/3 or x=1#