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anonymous
 one year ago
Solve 2x2 7x + 3 = 0.
anonymous
 one year ago
Solve 2x2 7x + 3 = 0.

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4\(\large\color{black}{ \displaystyle 2x^2 7x + 3 = 0 }\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what can i multiply to get 3 that i can add to get 7...nothing right?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4the discriminant is \(\large\color{black}{ \displaystyle (b)^2~4~\times~ (a)~\times~(c) }\) \(\large\color{black}{ \displaystyle (7)^2~4~\times~ (2)~\times~(3) }\) \(\large\color{black}{ \displaystyle 49~24 }\) \(\large\color{black}{ \displaystyle 25 }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4the discriminant is a positive integer and that means that you can factor your equation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so i use the formula \[b \pm \sqrt{b4(a)(c)/2}\]

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4yes, you can do that too \(\large\color{black}{ \displaystyle {\rm x}=\frac{b\pm\sqrt{b4(a)(c) \color{white}{\LARGE } }}{2a} }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4i don't think it is correct

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4also, wouldn't factoring be a more efficient method than quadratic formula ?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4but, if you want to use the quadratic formula, then sure, go ahead...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i tried factoring, and what i am being taught, is teaching the quadraic formula

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4ok, what I will suggest is to do it both methods.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0show me how to factor it

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4Have you ever factored an equation before?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{blue}{\text{Originally Posted by}}\) @jordanawest22 what can i multiply to get 3 that i can add to get 7...nothing right? \(\color{blue}{\text{End of Quote}}\) leading coefficient is not one you need to find two number if you multiply you should product of AC that's why...you can't find any number to get 7

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4how did you do it. was it like you were sort of "guessing" ?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4if so, then we can start doing the same thing.... it would be hard

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im doing \[2x ^{2}  7x 3...\right? so now i need \to split up 7 ...oh i did the oppsite my mistake\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what can i multiply to get 3 that i can add to get 7, the only numbers that i can multiply to get 3 is, 1, and 3...and1, and 3, but no matter what i do theydont make 7??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0should i divide all by 2?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4\(\large\color{black}{ \displaystyle 2x^27x+3=0 }\) at first we will write the parenthesis, and then we will start filling everything in as we go. \(\large\color{black}{ \displaystyle (~~~~~~~~~~~)(~~~~~~~~~~)=0 }\) you have to have \(\rm 2x^2\) as your first term and that you can get from a product of \(\rm x\cdot 2x\) So here, come the first two terms \(\large\color{black}{ \displaystyle (2x~~~~~~~~~)(x~~~~~~~~~)=0 }\) right now when we expand, we get: \(\large\color{black}{ \displaystyle 2x^2+?+?=0 }\) we don't know (yet) what we get for the rest of the terms, but the first term is going to be \(2x^2\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4(i am writing all the thinking behind it, but once you practice more, and perhaps if you watch some videos about it, you will find it easy. )

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what can i multiply to get 3 that i can add to get 7, the only numbers that i can multiply to get 3 is, 1, and 3...and1, and 3, but no matter what i do theydont make 7??

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4i don't really get what you are trying to do, but even though I don't understand it, it looks incorrect.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im trying to do....\[2x ^{2} ?\pm?+ 3....\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0onc ei get that then i can do (2x )(x )

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and inorder to get thos 2 numbers i need to no the numbers that multiply to get 3 that can add to get to 7

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4if you go that way, then \(\large\color{black}{ \displaystyle 2x^27x+3=0 }\) \(\large\color{black}{ \displaystyle 2x^24x3x+3=0 }\) \(\large\color{black}{ \displaystyle \color{red}{2x^24x}\color{blue}{3x+3}=0 }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4and you can factor the part in red and in blue separately (i labeled just for vision, not because it is a math denotation) and then factor it all togther

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for example \[x ^{2} + 4x +4.... would then be x^2 +2x +2x + 4\]

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4yes, I did that above

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but 4x, times 3x doesnt equal 3?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4i rewrote 7x as 4x3x

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4I am just wondering who taught you .... you are going from some unknown direction. I really don't get what you are doing here from a mathematical standpoint. this should help https://www.khanacademy.org/math/algebrabasics/corealgebraexpressions/corealgebramanipulatingexpressions/v/factoringalgebraicexpressions

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{blue}{\text{Originally Posted by}}\) @jordanawest22 what can i multiply to get 3 that i can add to get 7, the only numbers that i can multiply to get 3 is, 1, and 3...and1, and 3, but no matter what i do theydont make 7?? \(\color{blue}{\text{End of Quote}}\) i guess the question u did bfre this one are like x^2 + 3x +2 find two number if you multiply them you should get 2 and if you add or subtract them you should get middle term (here leading coefficient is one so AC = 1 times 2=2 right but in this case LEADING COEFFICIENT isn't one \[\huge\rm Ax^2+Bx+C=0\] you should multiply leading coefficient by constant term so \[\large\rm \color{reD}{ 2}x^2 7x + \color{reD}{3} = 0\] multiply A times C a=2 c =3 so ac = 6 NOW find two numbers if you multiply them you should get product of AC which is 6 and if you add or subtract them you should get middle term which is 7

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.02 numbers are 6 times 1 = 6 61 = 7 now remember LEADing coefficient is not one so you can't write (x + 1st number)(x+2nd number ) dw:1433898717154:dw carry down first and last term and then group method! so ur method is right but you don't know that you have to multiply A times C
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