How do you solve #(x)(x+1)=56#?

2 Answers
George C.
Aug 5, 2017

#x = 7" "# or #" "x = -8#

Explanation:

Given:

#(x)(x+1) = 56#

Here's one approach...

We want a pair of factors of #56# which differ by #1#.

Note that #7*8 = 56#, so one solution is #x=7#

The given equation is a quadratic (since if multiplied out it has term #x^2# of degree #2#), so will normally have two solutions. So what is the other?

The product of two negative numbers is positive, so we find the other solution given by:

#(-8)(-7) = 56#

That is: #x=-8#

Aswin K.
Aug 5, 2017

#x = 7# # &# # -8#

Explanation:

#(x)(x+1) = 56#

=> #x²+x = 56#

=> #x²+x-56 = 0#

=> Apply Quadratic Formula, #x# = #(-b+-sqrt(b^2-4*a*c))/(2*a)#

=> # a = 1, b = 1, c = -56 #

=> #x# = #(-1+-sqrt(1^2-(4*1*-56)))/(2*1)#

=> #x# = #(-1+-sqrt(1-(-224)))/2#

=> #x# = #(-1+-sqrt(1+224))/2#

=> #x# = #(-1+-sqrt225)/2#

=> #x# = #(-1+-15)/2#

=> #x# = #(-1+15)/2# #&# #(-1-15)/2#

=> #x# = #14/2# #&# #-16/2#

=> #x = 7# # &# # -8#

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