## Sunshine447 3 years ago 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

1. Sunshine447

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!

2. jiteshmeghwal9

sorry dude

3. Sunshine447

why?

4. jiteshmeghwal9

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

5. Sunshine447

LHS?

6. ParthKohli

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

7. Sunshine447

8. Yahoo!

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

9. ParthKohli

All right then. Do you know what the standard form of a quadratic equation is?

10. Yahoo!

@Sunshine447 Did u understand!

11. Sunshine447

@ParthKohli yes

12. ParthKohli

Okay, what is it?

13. Sunshine447

ax + bx + c = 0 right?

14. ParthKohli

\(ax^2 + bx + c = 0\)* What you first do is find the product of \(a\) and \(c\). What are \(a\) and \(c\) here?

15. Yahoo!

@Sunshine447 Did u understand my method????

16. Sunshine447

a=1 and c=12

17. ParthKohli

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

18. Sunshine447

12

19. ParthKohli

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

20. ParthKohli

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

21. Sunshine447

no

22. ParthKohli

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.

23. Sunshine447

-3 and -4

24. ParthKohli

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

25. ParthKohli

Do you get it till this point?

26. Sunshine447

yeah i think so

27. ParthKohli

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?

28. Sunshine447

yeah

29. ParthKohli

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

30. ParthKohli

Are you there? Do you get it?

31. Sunshine447

yeah

32. ParthKohli

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

33. Sunshine447

i dont think so

34. ParthKohli

All right, you have two solutions here. \( \color{Black}{\Rightarrow x - 4 = 0}\) \( \color{Black}{\Rightarrow x - 3 = 0}\)

35. Sunshine447

okay...

36. ParthKohli

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

37. Sunshine447

x=4 and 3

38. Sunshine447

Thanks!

39. ParthKohli

You're welcome!

40. TheViper

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

41. TheViper

So A. is the right answer ;)

42. TheViper

Understood ???