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

1. anonymous

that was 2x^2

2. SolomonZelman

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

3. anonymous

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

4. SolomonZelman

the discriminant is $$\large\color{black}{ \displaystyle (b)^2~-4~\times~ (a)~\times~(c) }$$ $$\large\color{black}{ \displaystyle (-7)^2~-4~\times~ (2)~\times~(3) }$$ $$\large\color{black}{ \displaystyle 49~-24 }$$ $$\large\color{black}{ \displaystyle 25 }$$

5. SolomonZelman

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

6. anonymous

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

7. SolomonZelman

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

8. anonymous

-7/4 +or- 25/4?

9. anonymous

hmmm

10. SolomonZelman

i don't think it is correct

11. anonymous

let me redo

12. SolomonZelman

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

13. SolomonZelman

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

14. anonymous

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

15. SolomonZelman

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

16. anonymous

show me how to factor it

17. SolomonZelman

ok.

18. SolomonZelman

Have you ever factored an equation before?

19. anonymous

yes

20. 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

21. SolomonZelman

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

22. SolomonZelman

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

23. anonymous

?

24. anonymous

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

25. anonymous

wait no i didnt

26. 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??

27. anonymous

should i divide all by 2?

28. 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 $$2x^2$$

29. 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. )

30. 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??

31. SolomonZelman

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

32. anonymous

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

33. anonymous

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

34. 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

35. SolomonZelman

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

36. 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

37. anonymous

for example $x ^{2} + 4x +4.... would then be x^2 +2x +2x + 4$

38. SolomonZelman

yes, I did that above

39. anonymous

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

40. SolomonZelman

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

41. 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

42. 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

43. 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