How do you solve #(x-3)(x+5)=x^2-7x-15#?

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Answer 1 · 2015-12-27T14:50:06

#color(blue)(x=0#

#color(blue)((x-3)(x+5) )= x^2-7x-15#

As per property :

#color(blue)((x+a)(x+b) = x^2 + (a+b)x +ab#

Applying the same to the L.H.S of the expression ,
where:# a=-3# and #b =5#:

#color(blue)((x-3)(x+5) ) = x^2 + (-3 +5) x + (-3) xx5#

#=color(blue)(x^2 +2x -15#

The expression now becomes:

#color(blue)(x^2 +2x -15)= x^2-7x-15#

#cancel(color(blue)(x^2)) +2x -cancel15= cancelx^2-7x-cancel15#

#2x=-7x#

#2x +7x =0#

#9x =0#

#x =0/9#

#color(blue)(x=0#