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LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.1Try applying the quadratic equation to this one.

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.1\[x = \frac{ b \pm \sqrt{b^{2}  4ac} }{ 2a }\]

Firejay5
 2 years ago
Best ResponseYou've already chosen the best response.1@BubbaMurphy Do you have a different way to do this problem???

BubbaMurphy
 2 years ago
Best ResponseYou've already chosen the best response.0LogicApple's way is definitely the way to do it.

Firejay5
 2 years ago
Best ResponseYou've already chosen the best response.1Is that really how yo would solve it

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.1Based on the equation you supplied, a = 1, b = b, and c = c.

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.1They are coefficients of whatever quadratic you are talking about. Did you have a quadratic in mind?

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.1\[6x^{2}  9x + 12 = 0\] \[5x^{2} + x  10 = 0\] \[x^2 + x  1 = 0\] These are some examples of quadratics, all set equal to 0.

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.1Oh, ok... so what is your question?

Firejay5
 2 years ago
Best ResponseYou've already chosen the best response.1We have to solve x^2 + bx + c = 0 for bonus and I have no idea what to do

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.1I suggested using the quadratic equation. The only difference between x^2 + bx + c = 0 and the quadratics that I mentioned is that the coefficients in your equation are only known by their variable letter. But that is ok, you can still apply the quadratic formula to them. The solutions to x^2 + bx + c = 0 is: \[\frac{ b \pm \sqrt{b^{2}  4c} }{ 2 }\] This is the same thing I wrote before, except I let a = 1 because the coefficient of x^2 in your equation is 1.

Firejay5
 2 years ago
Best ResponseYou've already chosen the best response.1We didn't learn that equation you gave me

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.1What about completing the square ?

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, but you would arrive at the same answer if you complete the square. Both techniques are valid for solving quadratic equations.

Firejay5
 2 years ago
Best ResponseYou've already chosen the best response.1I have to show my work, so how would I show it

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.1Well if you're using the quadratic equation you could say "applying the quadratic equation". But if you wanted to complete the square then: dw:1359081924838:dw

Azteck
 2 years ago
Best ResponseYou've already chosen the best response.0I think you can use sum and product of roots: \[\alpha +\beta= \frac{ b }{ a }\] \[\alpha \beta=\frac{ c }{ a }\]

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1359082578453:dw

Firejay5
 2 years ago
Best ResponseYou've already chosen the best response.1After (x + b/2)^2 = b^2/4  c. Find the square root of both.

Firejay5
 2 years ago
Best ResponseYou've already chosen the best response.1I got x + b/2 = b/4  c. Is that correct?

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.1no the square root of both sides gives: x + b/2 = ± sqrt(b^2/4  c)

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.1All the steps are in the picture attached.

Firejay5
 2 years ago
Best ResponseYou've already chosen the best response.1I am sorry the last step in attachment is the answer

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.1Oh ok did you get it figured out ?

Firejay5
 2 years ago
Best ResponseYou've already chosen the best response.1so x = +/ sqrt b^2  4c /2 is the answer

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1359083804186:dw yes

Firejay5
 2 years ago
Best ResponseYou've already chosen the best response.1Will you forgive me @LogicalApple ?

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.1What for? It's a learning experience

Firejay5
 2 years ago
Best ResponseYou've already chosen the best response.1I probably annoyed you a lot
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