Question

Solve the equation by factoring:
x2 - 7x + 12 = 0


A)
x = 3 or 4


B)
x = -3 or 4


C)
x = 3 or -4


D)
x = -3 or -4

Answer 1

please don't just give me the answer.... I want to actually try to learn how to do it. Also, know how to do this the "plug and play" method, but I need to know how to do it when it's not multiple choice. Thanks!

Answer 2

sorry dude

Answer 3

why?

Answer 4

first factorize the L.H.S. & then gt ur answer

Answer 5

LHS?

Answer 6

Hi, Sunshine! Would you like to know the whole process from the start?

Answer 7

yes please! Thanks Spiderman! haha

Answer 8

Sum = -7

Product = 12

two numbers -4 and -3


x^2 -4x - 3x + 12 =0


x(x-4) -3(x-4)


(x-4) (x-3)


x = 3 or x = 4

Answer 9

All right then.

Do you know what the standard form of a quadratic equation is?

Answer 10

@Sunshine447 Did u understand!

Answer 11

@ParthKohli yes

Answer 12

Okay, what is it?

Answer 13

ax + bx + c = 0 right?

Answer 14

\(ax^2 + bx + c = 0\)*

What you first do is find the product of \(a\) and \(c\).
What are \(a\) and \(c\) here?

Answer 15

@Sunshine447 Did u understand my method????

Answer 16

a=1 and c=12

Answer 17

Right! What's the product of 1, 12?

Answer 18

12

Answer 19

Correct! :)
Now you have to find two numbers that add to get \(b\) and multiply to get the product of \(a\) and \(c\).

Answer 20

Do you know two numbers such that:
Their sum is -7.
Their product is 12.
?

Answer 21

no

Answer 22

Hmm, you may find them by trial and error.
Note that a negative number times a negative number is a positive number, so those numbers must be negative if the sum of negative.

Answer 23

-3 and -4

Answer 24

You got it!
Now, we will 'split' the middle term with these coefficients.
\(-7x = -3x - 4x \Longrightarrow x^2 - 7x + 12 = x^2 - 3x - 4x + 12\)

Answer 25

Do you get it till this point?

Answer 26

yeah i think so

Answer 27

Okay, now we basically factor by 'grouping'.
What you must do is factor the first two and last two terms.
\(x^2 - 3x = x(x - 3)\)
\(-4x + 12 = -4(x - 3)\)
Get it till here?

Answer 28

yeah

Answer 29

So, \(x^2 - 3x - 4x + 12 = x(x - 3) - 4(x - 3)\).

\(x(x - 3) - 4(x - 3) \Longrightarrow ( x - 4)( x - 3)\)

Answer 30

Are you there? Do you get it?

Answer 31

yeah

Answer 32

Okay, so \((x - 4)(x - 3) = 0\).
Do you know the 'zero-product rule'?

Answer 33

i dont think so

Answer 34

All right, you have two solutions here.

\( \color{Black}{\Rightarrow x - 4 = 0}\)

\( \color{Black}{\Rightarrow x - 3 = 0}\)

Answer 35

okay...

Answer 36

Can you solve those equations?
That is how you get the solutions.

Answer 37

x=4 and 3

Answer 38

Thanks!

Answer 39

You're welcome!

Answer 40

\[\LARGE{x^2-7x+12=0}\]\[\Large{x^2-3x-4x+12=0}\]\[\Large{x(x-3)-4(x-3)=0}\]\[\Large{(x-3)(x-4)}\]\[\Large{\color{green}{x=3\space or \space4}}\]

Answer 41

So A. is the right answer ;)

Answer 42

Understood ???