How do you find the roots of #x^2+x=56#?

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Answer 1 · 2016-05-14T19:50:35

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

Note that #x^2+x = x(x+1)# and #56 = 7 * 8#

So #x = 7# is a root.

This is a quadratic, so it has another root.

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

So #x = -8# is the other root.

#color(white)()#
Alternative method

Alternatively, we can subtract #56# from both sides to get:

#x^2+x-56 = 0#

This is then in the form #ax^2+bx+c = 0# with #a=1#, #b=1# and #c = -56#.

This has roots given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

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

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

#=(-1+-sqrt(225))/2#

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

We find:

#(-1+15)/2 = 14/2 = 7#

#(-1-15)/2 = (-16)/2 = -8#