## Janu16 one year ago Using the completing-the-square method, rewrite f(x) = x2 − 8x + 3 in vertex form.

1. SolomonZelman

Ok, tell me this: Say, you have: $$x^2-8x$$ THEN, what number would you want to add to the end, to make that a perfect square trinomial?

2. Janu16

8? I think

3. Janu16

probably not

4. SolomonZelman

$$\large\color{black}{ \displaystyle x^2\pm\color{red}{\rm 2a}x+\color{red}{\rm a}^2 }$$

5. SolomonZelman

Does this look familiar to anything?

6. SolomonZelman

(x - a)² = x² - 2a + a² (x + a)² = x² + 2a + a²

7. Janu16

umm no. I learned this but can't get the answer

8. Janu16

wait i know that

9. SolomonZelman

you know those two in gray, right?

10. Janu16

sum and difference of a cube??

11. SolomonZelman

No cubes: Just how to expand (x+a)^2 and (x-a)^2

12. Janu16

ohok

13. SolomonZelman

So, we are going to make x^2-8x (by adding some number) into x^2-2a+a^2 form

14. Janu16

ok

15. SolomonZelman

$$\large\color{black}{ \displaystyle x^2-\color{red}{\rm 2a}x+\color{red}{\rm a}^2 =(x-\color{red}{\rm a})^2 }$$

16. SolomonZelman

So if 2a is 8, than a is 4. Right?

17. Janu16

ya

18. Janu16

than you square it?

19. SolomonZelman

$$\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x+\color{red}{\rm 4}^2 =(x-\color{red}{\rm 8})^2 }$$ $$\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x+16 =(x-\color{red}{\rm 8})^2 }$$ that would be a perfect square trinomial. So we add 16 to the x^2-8x, to make our polynomial a perfect square.

20. Janu16

what is this formula called? cause I didn't learn this?

21. SolomonZelman

which formula?

22. Janu16

the one you are using

23. SolomonZelman

oh, don't worry, you can just google (a+b)^2 and all of that will come up

24. SolomonZelman

all of those like (a+b)^2 and (a-b)^2 ... it will all show up.

25. Janu16

okk

26. SolomonZelman

We will add 16, but we can NOT just add 16, because we will change the value of the trinomial. So this is what we will do: $$\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x\color{blue}{}+3 }$$ $$\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x\color{blue}{+16-16}+3 }$$ $$\large\color{black}{ \displaystyle (x^2-\color{red}{\rm 8}x+16)-16+3 }$$ in parenthesis, it is a perfect square trinomial and -16+3 is -13.

27. Janu16

so it would be -8-13+3

28. SolomonZelman

you mixed up something a little, how did you get that result/

29. SolomonZelman

??

30. Janu16

from that equation

31. Janu16

wait x- 8 - 13?

32. SolomonZelman

$$\large\color{black}{ \displaystyle x^2-\color{red}{\rm 2a}x+\color{red}{\rm a}^2 =(x-\color{red}{\rm a})^2 }$$ And we want to get our $$x62-8x$$ into that form, by adding a value. We will add 16, but we can NOT just add 16, because we will change the value of the trinomial. So this is what we will do: $$\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x\color{blue}{}+3 }$$ $$\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x\color{blue}{+16-16}+3 }$$ $$\large\color{black}{ \displaystyle (x^2-\color{red}{\rm 8}x+16)-16+3 }$$ I added a magic zero, so to speak. +16-16, is all I did:) in parenthesis, it is a perfect square trinomial and -16+3 is -13. $$\large\color{black}{ \displaystyle (x^2{~}-{~} 2\times\color{red}{ 4}x{~}+{~}\color{red}{ 4}^2)-13 }$$ do you see that x$$^2$$-2a+a$$^2$$ form in the parenthesis? We know that a²-2a+a²=(x-a)², and thus we get the following: $$\large\color{black}{ \displaystyle (x-\color{red}{ 4})^2-13 }$$

33. SolomonZelman

34. Janu16

ohok tysm!! so you just simplied 8 to 4?

35. SolomonZelman

oh, the result I showed at the end can't be simplified. You can re-write it to what it was in the beginning if you do the inverse operations.

36. Janu16

ohhhok f(x) = (x − 4)2 − 13 this would be the answer right?

37. SolomonZelman

(x-4)$$\color{red}{^2}$$-13 :D

38. Janu16

tysm! can you hep with other one on new thread plz?