How do you solve #-6x + 9 = -x^2#?

Question

357 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-07T13:37:27

See a solution process below:

First, add #color(red)(x^2)# to each side of the equation to put the equation in standard quadratic form:

#color(red)(x^2) - 6x + 9 = color(red)(x^2) - x^2#

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

We can now factor the left side of the equation as;

#(x - 3)(x - 3) = 0#

#(x - 3)^2 = 0#

Because both factors are the same there is one solution for this problem. We can equate the term to #0# to solve for #x#;

#x - 3 = 0#

#x - 3 + color(red)(3) = 0 + color(red)(3)#

#x - 0 = 3#

#x = 3#

Answer 2 · 2018-02-07T13:38:43

#x=3#

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

Or:

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

This is a quadratic equation of the form #ax^2+bx+c=0#, where #x=(-b+-sqrt(b^2-4ac))/(2a)#.

Here, #a=1, b=-6, c=9#.

So input:

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

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

#(6+-sqrt(0))/2#

#(6+-0)/2#.

Since #x+0=x-0#, this becomes:

#6/2#

#x=3#