Question

What are the solutions of the quadratic equation?
x^2 - 9x + 20 = 0

Answer 1

In this problem, we can observe it's a 2nd degree equality, because the "x" with the highest exponent is the "x2 with a exponent of 2.
So, we have to use the general formula:
\[x=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]
all you have to do is replace the values :)

Answer 2

thanks!

Answer 3

Can you stick around until I get my answer?

Answer 4

@Owlcoffee
Im stuck I got to
\[x = -9 \pm \sqrt{1}\]

Answer 5

Oh, you made a little mistake :(

Answer 6

nooo. help :(

Answer 7

So, any quadratic equation has a general form of:
\[ax^2 + bx + c=0\]
in order to find the value of x, the general formula comes in action:
\[x=\frac{ -b \pm \sqrt{b^2 -4ac} }{ 2a }\]
so, in the problem:
\[x^2 -9x +20=0\]
as a trick I used back in the day, I separeted them as a sum, so I could see a,b and c more clearly:
\[x^2 + (-9)x+20=0\]
so, then:
\[a=1 , b=-9,c=20\]
recplacing in the quadratic formula:
\[x=\frac{ -(-9) \pm \sqrt{(-9)^2 -4(1)(20)} }{ 2(1) }\]
can you take over from here? :)

Answer 8

well thats what I did! and I simplified and ended up with \[x = 9 \pm \sqrt{1}\] all over 2

Answer 9

@Owlcoffee

Answer 10

and I got stuck there.

Answer 11

good, and what is the square root of 1?

Answer 12

1

Answer 13

good, so we have:
\[x=\frac{ 9 \pm 1 }{ 2 }\]
so, in order to get rid of the plus/minus, we have to take both situations:
\[x_1 = \frac{ 9 +1 }{ 2 }\]
\[x_2 = \frac{ 9-1 }{ 2 }\]