how would I solve a negative quadratic function?
example: f(x) = -x^2+6x-5

Question

how would I solve a negative quadratic function?
example: f(x) = -x^2+6x-5

anonymous · Answer

$$ -x^2+6x-5=0\iff x^2-6x+5=0$$ solve the "positive" one

anonymous · Answer

same way you would a positive one. only A=-1 insead of 1. or you could just multiply both sides by -1 to make the leading coefficient to positive.

hero · Answer

First, set f(x) = 0:
0 = -x^2 + 6x - 5

Next factor out the negative:
0 = -(x^2 - 6x + 5)

Then divide both sides by -1:
0 = x^2 - 6x + 5

hero · Answer

@Satellite73 is such a sniper.

anonymous · Answer

yep

anonymous · Answer

(x-5)(x-1)
divide both by -1 = x=-5 x=-1?

hero · Answer

After dividing both sides by -1, you don't have to do that process anymore.

anonymous · Answer

your factoring is right. just dont re divide.

hero · Answer

You only do it once, then solve the quadratic as you normally do.

anonymous · Answer

if you did redivide it would be -x=-5 though keep that in mind

hero · Answer

(x - 5)(x - 1) = 0

x - 5 = 0
x - 1 = 0

Then solve from there.

anonymous · Answer

ohh i see thank you guys :D