anonymous
  • anonymous
Solve x2- 5x + 6 = 0 using the quadratic formula.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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pooja195
  • pooja195
List your abc values
Nnesha
  • 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
  • anonymous
?

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nuttyliaczar
  • nuttyliaczar
Imagine your equation is in the form Ax^2+Bx+C=0. What is your A, B, and C?
pooja195
  • pooja195
What are the A B C values
nuttyliaczar
  • 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
  • anonymous
a=1 b=5 and c=6
pooja195
  • pooja195
hmm a=1 b= -5 c=6
anonymous
  • anonymous
cant i do x^2 +2x + 3x -5?
pooja195
  • pooja195
LEts stick to this its less complicated
pooja195
  • pooja195
\[\huge~x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a } \]
pooja195
  • pooja195
fimiliar with this?
anonymous
  • anonymous
x2- 5x + 6 = 0 x^2 - 2 + 3 + 6 (x^2-2x) (3x+6) divide by x and 2 so....
anonymous
  • anonymous
yes
pooja195
  • pooja195
By factoring? Its asking to use the quadratic formula .
pooja195
  • pooja195
\[\huge~x=\frac{ -(-5) \pm \sqrt{(-5)^2-4(1)(6)} }{ 2(1) } \]
pooja195
  • pooja195
simplyfy this
anonymous
  • 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
  • anonymous
\[5\pm/2 \sqrt{-6}\]
anonymous
  • anonymous
Oh. I got all the numbers wrong...
anonymous
  • anonymous
Very sorry
pooja195
  • pooja195
\[\frac{ 5\pm \sqrt{1} }{ 2 }\]
pooja195
  • pooja195
simplfy
pooja195
  • pooja195
Set up 2 equations
anonymous
  • anonymous
\[5\pm 1/2\]
anonymous
  • anonymous
oh
anonymous
  • anonymous
The equation is 5+-(5^2 -4(1)(6))^1/2 /2(1). ^1/2 means square root
anonymous
  • anonymous
jpro, can u stop talking
pooja195
  • 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
  • anonymous
so + that and - that
pooja195
  • pooja195
yea :)
anonymous
  • anonymous
is that the answer?
anonymous
  • anonymous
x = 2 and x = 3 x = - 4 and x = - 3 x = 5 and x = - 3 x = - 2 and x = 4
anonymous
  • anonymous
3 and 2
pooja195
  • pooja195
yep :)
pooja195
  • pooja195
Thats right good job :)
anonymous
  • anonymous
i learned this along time ago the fist thing i learned, but what im doing now has nothing to with it...
pooja195
  • pooja195
lol what are you learning right now ?
anonymous
  • anonymous
things like x^2 +6=180
pooja195
  • pooja195
Oh so like factoring?
pooja195
  • pooja195
Let me know if you need any help :)
pooja195
  • pooja195
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anonymous
  • anonymous
not really were r kinda using a diffent methon with squareroots
pooja195
  • pooja195
oh right thats what its called :)

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