How do you solve #x^2 -x = 5x - 9#?

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Answer 1 · 2015-07-10T09:03:43

#x=3#

#x^2-x=5x-9#

Subtract #5x# from both sides of the equation.

#x^2-x-5x=-9# =

#x^2-6x=-9#

Add #9# to both sides.

#x^2-6x+9=0#

We now have a quadratic equation with the form #ax^2+bx+c=0#, where #a=1;# #b=-6;# and #c=9#.

We can solve for #x# using the quadratic formula.

Quadratic Formula

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

#x=(-(-6)+-sqrt(-6^2-4*1*9))/(2*1)# =

#x=(6+-sqrt(36-36))/2# =

#x=(6+-sqrt0)/2# =

#x=(6+-0)/2# =

#x=6/2# =

#x=3#