Solve x2- 5x + 6 = 0 using the quadratic formula.

- anonymous

Solve x2- 5x + 6 = 0 using the quadratic formula.

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- pooja195

List your abc values

- Nnesha

\[\huge\rm \frac{ - b \pm \sqrt{b^2 -4ac} }{ 2a }\]
quadratic formula
where abc values are \[\huge\rm Ax^2 +Bx+C=0\]

- anonymous

?

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## More answers

- nuttyliaczar

Imagine your equation is in the form Ax^2+Bx+C=0. What is your A, B, and C?

- pooja195

What are the A B C values

- nuttyliaczar

Also sometimes you would have to manipulate your equation to get it in that form but in this case it's set up nicely for you

- anonymous

a=1
b=5
and
c=6

- pooja195

hmm
a=1
b= -5
c=6

- anonymous

cant i do x^2 +2x + 3x -5?

- pooja195

LEts stick to this its less complicated

- pooja195

\[\huge~x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a } \]

- pooja195

fimiliar with this?

- anonymous

x2- 5x + 6 = 0
x^2 - 2 + 3 + 6
(x^2-2x) (3x+6)
divide by x and 2
so....

- anonymous

yes

- pooja195

By factoring? Its asking to use the quadratic formula .

- pooja195

\[\huge~x=\frac{ -(-5) \pm \sqrt{(-5)^2-4(1)(6)} }{ 2(1) } \]

- pooja195

simplyfy this

- anonymous

So the x2 can be written as 2x because they are communitive terms. Since the 2x and 5x are like terms, they can combine so 2x-5x = -3x. Ok. Now, since u want to use quadratic u need to assign an a b and c value. The a value is the coefficient of an x^2 in a quadratic equation. Since u don't have a quadratic equation, the coefficient is 0. So the a value is zero. The b value is the -3 and the c value is the number without an x (I don't remember what it was.) Plug the a, b and c values into the quadratic equation and solve. I hope that helps, because i am not good at explaining things. If u still don't understand, I'm sure Kahn academy has a video on it

- anonymous

\[5\pm/2 \sqrt{-6}\]

- anonymous

Oh. I got all the numbers wrong...

- anonymous

Very sorry

- pooja195

\[\frac{ 5\pm \sqrt{1} }{ 2 }\]

- pooja195

simplfy

- pooja195

Set up 2 equations

- anonymous

\[5\pm 1/2\]

- anonymous

oh

- anonymous

The equation is 5+-(5^2 -4(1)(6))^1/2 /2(1). ^1/2 means square root

- anonymous

jpro, can u stop talking

- pooja195

thats what i have above xD
\[\frac{ 5+\sqrt{1} }{ 2}\]
\[\frac{ 5-\sqrt{1} }{ 2}\]
I did this because we have a plus and minus sign so we need 2 seperate equations

- anonymous

so + that and - that

- pooja195

yea :)

- anonymous

is that the answer?

- anonymous

x = 2 and x = 3
x = - 4 and x = - 3
x = 5 and x = - 3
x = - 2 and x = 4

- anonymous

3 and 2

- pooja195

yep :)

- pooja195

Thats right good job :)

- anonymous

i learned this along time ago the fist thing i learned, but what im doing now has nothing to with it...

- pooja195

lol what are you learning right now ?

- anonymous

things like x^2 +6=180

- pooja195

Oh so like factoring?

- pooja195

Let me know if you need any help :)

- pooja195

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- anonymous

not really
were r kinda using a diffent methon with squareroots

- pooja195

oh right thats what its called :)

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