How do you solve #9=x+x^2#?

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Answer 1 · 2018-06-09T17:21:38

#x= (-1+-sqrt(37))/2#

#9=x+x^2#

You can complete the square:

#ax^2+bx+c#

a must equal 1

#c=(b/2)^2#

square term #=b/2#

#x^2+x = 9#

#a=1# so we can proceed.

#c=(1/2)^2=1/4#

#x^2+x +c = 9+c#

#x^2+x +1/4 = 9+1/4#

factor the left side:

#(x+1/2)(x+1/2) = 9+1/4#

#(x+1/2)^2 = 37/4#

now solve:

#sqrt((x+1/2)^2) = +-sqrt(37/4)#

#x+1/2= +-sqrt(37/4)#

#x= -1/2+-sqrt(37)/2#

#x= (-1+-sqrt(37))/2#

graph{x^2+x -9 [-20, 20, -10, 10]}