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anonymous

  • one year ago

Solve 2x2 -7x + 3 = 0.

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  1. anonymous
    • one year ago
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    that was 2x^2

  2. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle 2x^2 -7x + 3 = 0 }\)

  3. anonymous
    • one year ago
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    what can i multiply to get 3 that i can add to get 7...nothing right?

  4. SolomonZelman
    • one year ago
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    the discriminant is \(\large\color{black}{ \displaystyle (b)^2~-4~\times~ (a)~\times~(c) }\) \(\large\color{black}{ \displaystyle (-7)^2~-4~\times~ (2)~\times~(3) }\) \(\large\color{black}{ \displaystyle 49~-24 }\) \(\large\color{black}{ \displaystyle 25 }\)

  5. SolomonZelman
    • one year ago
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    the discriminant is a positive integer and that means that you can factor your equation.

  6. anonymous
    • one year ago
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    so i use the formula \[-b \pm \sqrt{b-4(a)(c)/2}\]

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

  8. anonymous
    • one year ago
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    -7/4 +or- 25/4?

  9. anonymous
    • one year ago
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    hmmm

  10. SolomonZelman
    • one year ago
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    i don't think it is correct

  11. anonymous
    • one year ago
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    let me redo

  12. SolomonZelman
    • one year ago
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    also, wouldn't factoring be a more efficient method than quadratic formula ?

  13. SolomonZelman
    • one year ago
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    but, if you want to use the quadratic formula, then sure, go ahead...

  14. anonymous
    • one year ago
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    i tried factoring, and what i am being taught, is teaching the quadraic formula

  15. SolomonZelman
    • one year ago
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    ok, what I will suggest is to do it both methods.

  16. anonymous
    • one year ago
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    show me how to factor it

  17. SolomonZelman
    • one year ago
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    ok.

  18. SolomonZelman
    • one year ago
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    Have you ever factored an equation before?

  19. anonymous
    • one year ago
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    yes

  20. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @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

  21. SolomonZelman
    • one year ago
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    how did you do it. was it like you were sort of "guessing" ?

  22. SolomonZelman
    • one year ago
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    if so, then we can start doing the same thing.... it would be hard

  23. anonymous
    • one year ago
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    ?

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

  25. anonymous
    • one year ago
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    wait no i didnt

  26. anonymous
    • one year ago
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    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??

  27. anonymous
    • one year ago
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    should i divide all by 2?

  28. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle 2x^2-7x+3=0 }\) 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 \(2x^2\)

  29. SolomonZelman
    • one year ago
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    (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. )

  30. anonymous
    • one year ago
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    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??

  31. SolomonZelman
    • one year ago
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    i don't really get what you are trying to do, but even though I don't understand it, it looks incorrect.

  32. anonymous
    • one year ago
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    im trying to do....\[2x ^{2} -?\pm?+ 3....\]

  33. anonymous
    • one year ago
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    onc ei get that then i can do (2x )(x )

  34. anonymous
    • one year ago
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    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

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

  36. SolomonZelman
    • one year ago
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    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

  37. anonymous
    • one year ago
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    for example \[x ^{2} + 4x +4.... would then be x^2 +2x +2x + 4\]

  38. SolomonZelman
    • one year ago
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    yes, I did that above

  39. anonymous
    • one year ago
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    but -4x, times 3x doesnt equal 3?

  40. SolomonZelman
    • one year ago
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    i rewrote -7x as -4x-3x

  41. SolomonZelman
    • one year ago
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    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

  42. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @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

  43. Nnesha
    • one year ago
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    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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