+1005 Plugging it in to the quadratic formula
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
We have.
\(x = {10 \pm \sqrt{-10^2-4*9} \over 2} = {10 \pm \sqrt{64}\over 2} = {10 \pm 8}\over2\)
So 10 +/- 8 OVER 2 is 18/2 or 1.
Plugging into the equation
9^2 - 90 +9 = 0 = 81 + 9 - 90 = 0
YUP!
1^2 - 10 + 9 = 0 = 1 + 9 - 10 = 10-10 = 0.
YUP!
Great those are our roots.
A quadratic can only have two,
If you don't understand anything feel free to ask!
Solve the quadratic x^2 - 10x + 9 = 0.
Alternatively, this one is simple enought that you could just factor it.
What two numbers multiply together to get +9 and add together to get –10 ?
I wish they were all this easy. Of course, the numbers are –1 and –9
So, x2 – 10x + 9 = 0 factored is (x –1)(x – 9) = 0
Setting each component equal to zero: (x – 1) = 0
x = 1
(x – 9) = 0
x = 9
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