Solve 2x2 -7x + 3 = 0.

- anonymous

Solve 2x2 -7x + 3 = 0.

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- anonymous

that was 2x^2

- SolomonZelman

\(\large\color{black}{ \displaystyle 2x^2 -7x + 3 = 0 }\)

- anonymous

what can i multiply to get 3 that i can add to get 7...nothing right?

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## More answers

- SolomonZelman

the 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

the discriminant is a positive integer and that means that you can factor your equation.

- anonymous

so i use the formula \[-b \pm \sqrt{b-4(a)(c)/2}\]

- SolomonZelman

yes, you can do that too
\(\large\color{black}{ \displaystyle {\rm x}=\frac{-b\pm\sqrt{b-4(a)(c) \color{white}{\LARGE |} }}{2a} }\)

- anonymous

-7/4 +or- 25/4?

- anonymous

hmmm

- SolomonZelman

i don't think it is correct

- anonymous

let me redo

- SolomonZelman

also, wouldn't factoring be a more efficient method than quadratic formula ?

- SolomonZelman

but, if you want to use the quadratic formula, then sure, go ahead...

- anonymous

i tried factoring, and what i am being taught, is teaching the quadraic formula

- SolomonZelman

ok, what I will suggest is to do it both methods.

- anonymous

show me how to factor it

- SolomonZelman

ok.

- SolomonZelman

Have you ever factored an equation before?

- anonymous

yes

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @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

how did you do it.
was it like you were sort of "guessing" ?

- SolomonZelman

if so, then we can start doing the same thing.... it would be hard

- anonymous

?

- anonymous

im doing \[2x ^{2} - 7x- 3...\right? so now i need \to split up 7 ...oh i did the oppsite my mistake\]

- anonymous

wait no i didnt

- anonymous

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...and-1, and -3, but no matter what i do theydont make 7??

- anonymous

should i divide all by 2?

- SolomonZelman

\(\large\color{black}{ \displaystyle 2x^2-7x+3=0 }\)
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

(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

- SolomonZelman

i don't really get what you are trying to do, but even though I don't understand it, it looks incorrect.

- anonymous

im trying to do....\[2x ^{2} -?\pm?+ 3....\]

- anonymous

onc ei get that then i can do (2x )(x )

- anonymous

and 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

if you go that way, then
\(\large\color{black}{ \displaystyle 2x^2-7x+3=0 }\)
\(\large\color{black}{ \displaystyle 2x^2-4x-3x+3=0 }\)
\(\large\color{black}{ \displaystyle \color{red}{2x^2-4x}\color{blue}{-3x+3}=0 }\)

- SolomonZelman

and 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

for example
\[x ^{2} + 4x +4.... would then be x^2 +2x +2x + 4\]

- SolomonZelman

yes, I did that above

- anonymous

but -4x, times 3x doesnt equal 3?

- SolomonZelman

i rewrote -7x as -4x-3x

- SolomonZelman

I 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/algebra-basics/core-algebra-expressions/core-algebra-manipulating-expressions/v/factoring-algebraic-expressions

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @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...and-1, 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

2 numbers are
-6 times -1 = 6
-6-1 = -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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