SolutionS of x^2+12x+10=0
-1 and -11
How do you get -1 and -11?
Is it right?
No, since x=-1 and x=-11 being solutions =>x+1=0 and x+11=0 (x+1)(x+11) = x^2 + 11x + x + 11 = x^2 + 12x + 11, not x^2 + 12x + 10
I don't know then
The discriminant is 104, which is not a perfect square and thus indicates that there are two solutions that are irrational. I'll use quadratic formula since I already have the discriminant: \[ x = \frac{-b \pm \sqrt{D}}{2a} \] \[ x=\frac{-12 \pm \sqrt{104}}{2(1)}\\ = \frac{-12 \pm \sqrt{4\times26}}{2}\\ = \frac{-\cancel{12}6 \pm \cancel2\sqrt{26}}{\cancel2}\\ x= -6 \pm \sqrt{26} \]
That's what I got but I just take the real number
I see now, though it would be more accurate to use \(x=-6 \pm \sqrt{26}\) than approximating to nearest integer :)
yea ok
my bad
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