Solve 2x2 -7x + 3 = 0.

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Solve 2x2 -7x + 3 = 0.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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that was 2x^2
\(\large\color{black}{ \displaystyle 2x^2 -7x + 3 = 0 }\)
what can i multiply to get 3 that i can add to get 7...nothing right?

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the discriminant is \(\large\color{black}{ \displaystyle (b)^2~-4~\times~ (a)~\times~(c) }\) \(\large\color{black}{ \displaystyle (-7)^2~-4~\times~ (2)~\times~(3) }\) \(\large\color{black}{ \displaystyle 49~-24 }\) \(\large\color{black}{ \displaystyle 25 }\)
the discriminant is a positive integer and that means that you can factor your equation.
so i use the formula \[-b \pm \sqrt{b-4(a)(c)/2}\]
yes, you can do that too \(\large\color{black}{ \displaystyle {\rm x}=\frac{-b\pm\sqrt{b-4(a)(c) \color{white}{\LARGE |} }}{2a} }\)
-7/4 +or- 25/4?
hmmm
i don't think it is correct
let me redo
also, wouldn't factoring be a more efficient method than quadratic formula ?
but, if you want to use the quadratic formula, then sure, go ahead...
i tried factoring, and what i am being taught, is teaching the quadraic formula
ok, what I will suggest is to do it both methods.
show me how to factor it
ok.
Have you ever factored an equation before?
yes
\(\color{blue}{\text{Originally Posted by}}\) @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
how did you do it. was it like you were sort of "guessing" ?
if so, then we can start doing the same thing.... it would be hard
?
im doing \[2x ^{2} - 7x- 3...\right? so now i need \to split up 7 ...oh i did the oppsite my mistake\]
wait no i didnt
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??
should i divide all by 2?
\(\large\color{black}{ \displaystyle 2x^2-7x+3=0 }\) 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 \(2x^2\)
(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. )
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??
i don't really get what you are trying to do, but even though I don't understand it, it looks incorrect.
im trying to do....\[2x ^{2} -?\pm?+ 3....\]
onc ei get that then i can do (2x )(x )
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
if you go that way, then \(\large\color{black}{ \displaystyle 2x^2-7x+3=0 }\) \(\large\color{black}{ \displaystyle 2x^2-4x-3x+3=0 }\) \(\large\color{black}{ \displaystyle \color{red}{2x^2-4x}\color{blue}{-3x+3}=0 }\)
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
for example \[x ^{2} + 4x +4.... would then be x^2 +2x +2x + 4\]
yes, I did that above
but -4x, times 3x doesnt equal 3?
i rewrote -7x as -4x-3x
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
\(\color{blue}{\text{Originally Posted by}}\) @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
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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