How do you solve #(11x - 5) ^ { 2} - ( 10x - 1) ^ { 2} - ( 3x - 20) ( 7x + 10) = 124#?

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Answer 1 · 2018-06-12T02:02:24

#color(crimson)(x = 40/53 #

Follow the PEMDAS Rule to solve.

#(11x - 5)^2 - (10x-1)^2 - (3x-2)(7x+10) = 124#

#121x^2 - 110x + 25 - 100x^2 + 20x - 1 - 21x^2 - 30x + 14x + 20 = 124, color(purple)(" Removing Braces"#

#121x^2 - 100x^2 - 21x^2 - 110x + 20x - 30x + 14x + 25 - 1 + 20 = 124, color(purple)("Bringing like terms together"#

#0 x^2 - 106x + 44 = 124, color(purple)("Adding / Subtracting"#

#106x = - 124 +44 #

#106x = - 80 #

#color(crimson)(x = 80/106 " or " x = 40/53, color(purple)("Simplifying"#

Answer 2 · 2018-06-12T02:08:19

#x=-5#

We have

#(11x - 5) ^ { 2} - ( 10x - 1) ^ { 2} #
#= (11x-5-10x+1)(11x-5+10x-1)#
#= (x-4)(21x-6) = 21x^2-90x+24#

and

# ( 3x - 20) ( 7x + 10) = 21 x^2-110x-200#

Thus, the equation is

#(21x^2-90x+24)-(21 x^2-110x-200)=124 implies#

#20x+ 100 = 0 implies 20(x+5) = 0 implies#

#x = -5#