Question

Solve 2x2 -7x + 3 = 0.

Answer 1

that was 2x^2

Answer 2

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

Answer 3

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

Answer 4

the discriminant is

\(\large\color{black}{ \displaystyle (b)^2~-4~\times~ (a)~\times~(c) }\)

\(\large\color{black}{ \displaystyle (-7)^2~-4~\times~ (2)~\times~(3) }\)

\(\large\color{black}{ \displaystyle 49~-24 }\)

\(\large\color{black}{ \displaystyle 25 }\)

Answer 5

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

Answer 6

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

Answer 7

yes, you can do that too

\(\large\color{black}{ \displaystyle {\rm x}=\frac{-b\pm\sqrt{b-4(a)(c) \color{white}{\LARGE |} }}{2a} }\)

Answer 8

-7/4 +or- 25/4?

Answer 9

hmmm

Answer 10

i don't think it is correct

Answer 11

let me redo

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

show me how to factor it

Answer 17

ok.

Answer 18

Have you ever factored an equation before?

Answer 19

yes

Answer 20

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

Answer 21

how did you do it.

was it like you were sort of "guessing" ?

Answer 22

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

Answer 23

?

Answer 24

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

Answer 25

wait no i didnt

Answer 26

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??

Answer 27

should i divide all by 2?

Answer 28

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

Answer 29

(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. )

Answer 30

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??

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

if you go that way, then

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

\(\large\color{black}{ \displaystyle 2x^2-4x-3x+3=0 }\)

\(\large\color{black}{ \displaystyle \color{red}{2x^2-4x}\color{blue}{-3x+3}=0 }\)

Answer 36

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

Answer 37

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

Answer 38

yes, I did that above

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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 )
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