imqwerty one year ago real math ( ͡~ ͜ʖ ͡°)

1. imqwerty

2. anonymous

D

3. imqwerty

correct @mukushla :)

4. anonymous

nice question

5. anonymous

we need to find out that $$b^2=ac$$

6. imqwerty

yes correct

7. ganeshie8

*

8. Zale101

Multiply by the conjugate to both the numerator and the denominator. Foil the numerator and the denominator and you'll get the answer.

9. imqwerty

well done @Zale

10. imqwerty

i forgot to write 1 step in the solution so heres the new solution

11. Sachintha

@perl Can you explain it?

12. perl

yes

13. Sachintha

What I didn't get is how $b^2=ac$

14. perl

We want to force the entire expression to be rational, given a,b,c are integers. You have two possibilities, b^2 -ac = 0 or sqrt(2) b^2 -ac cannot equal to sqrt(2) , because of the denominator. So you are left with only b^2 -ac = 0

15. Sachintha

I still didn't get why $b^2-ac=0$ Is it taken so that $\sqrt2$ can be removed?

16. perl

otherwise you are left with a radical 2 in the numerator, so the entire expression is irrational

17. perl

yes to remove the square root 2

18. perl

do you see why b^2 -ac cannot equal to square root 2 ?

19. perl

edit** $\Large{ \frac{a\sqrt{2}+b}{b\sqrt{2} + c }\cdot \frac{b\sqrt{2}-c}{b\sqrt{2} -c } \\~\\= \frac{2ab + b^2\sqrt 2 - ac \sqrt 2 - bc}{2b^2 - c^2 } \\~\\= \frac{2ab + \sqrt 2 (b^2 - ac) - bc}{2b^2 - c^2 } }$

20. Sachintha

Ohh I forgot all about sets of real numbers including rational and irrational .. :D But what does number are a,ar and ar^2 mean?

21. perl

that part is a bit unclear. we can try to figure it out

22. Sachintha

I think, $c=ar^2$

23. Sachintha

r is not defined here :(

24. Sachintha

So b/a is taken as r ?

25. Sachintha

Is it possible to equalize it like that?

26. Sachintha

Well it does work too. :)

27. Sachintha

So we equalize it to another value for the ease of calculations? Sorry for being so dumb.

28. perl

these are good questions :)

29. perl

since b^2 = ac , divide both sides by a $$\Large c = \frac{b^2}{ a} = a \cdot \frac{b^2}{ a^2 } = a \cdot \left(\frac b a \right)^2$$ Let r = (b/a) Now we have a= a ar = a (b/a) = b ar^2= a (b/a)^2 = c

30. Sachintha

I got it. :D Was confused because of r.

31. perl

Yes it looked like it came from out of no where :D

32. perl

In part d) it appears they completed long division.

33. Sachintha

Did long division for myself. :D (whew) It factors to $(r^2-r+1)(r^2+r+1)$

34. perl

i see, great job

35. Sachintha

Thank you so much for taking your time and explaining me. :)

36. perl

why can't b^2 -ac = sqrt(2) ?

37. Sachintha

Because we still get a irrational number at the denominator. I am correct?

38. perl

yes can you show why

39. Sachintha

Because $b^2=\sqrt2+ac$

40. perl

right

41. Sachintha

Am I wrong?

42. Sachintha

43. perl

if we plug in the denominator we get 2b^2 -c^2 = 2 ( √2 + ac) - c^2 = 2√2 + 2ac - c^2 , which is irrational

44. perl

because of the 2 sqrt(2)

45. Sachintha

yep :)