## anonymous 2 years ago Show that kb^2=(k+1)^2*ac

1. anonymous

Hi :-) what is your constraint?

2. anonymous

One of the roots of the quadratic equation $ax ^{2}+bx+c=0$ is k times the other root. Show that $kb ^{2}=(k+1)^{2}ac$

3. anonymous

have you tried it yet?

4. anonymous

$\frac{ -b+\sqrt{b ^{2}-4ac} }{ 2a } = k \frac{ -b-\sqrt{b ^{2}-4ac} }{ 2a }$

5. anonymous

That's how I started, but I got stuck, and it gets kinda messy

6. anonymous

let's play with variable roots, assume that roots are $$x_1$$ and $$x_2$$

7. anonymous

what are the expressions for $$x_1 x_2$$ and $$x_1+x_2$$ in terms of quadratic coefficients?

8. anonymous

Not sure I understand what you mean? Two roots are in ratio 1:k, so x1=kx2

9. anonymous

I can get x1 and x2 by doing the -b formula, as I wrote above, one with + another with -. And than I multiply one of them with k.

10. anonymous

do u know about formula for the Sum and product of the roots of a quadratic?

11. anonymous

Don't think so

12. anonymous

I will teach you then :-)

13. anonymous

well as you know$x_1=\frac{ -b+\sqrt{b ^{2}-4ac} }{ 2a }$$x_2=\frac{ -b-\sqrt{b ^{2}-4ac} }{ 2a }$

14. anonymous

now let's calculate $$x_1+x_2$$: $x_1+x_2=\frac{ -b+\sqrt{b ^{2}-4ac} }{ 2a }+\frac{ -b-\sqrt{b ^{2}-4ac} }{ 2a }\\=\frac{ -b+\sqrt{b ^{2}-4ac} -b-\sqrt{b ^{2}-4ac} }{ 2a }=-\frac{b}{a}$

15. anonymous

Ok :) thats reasonably

16. anonymous

am i clear nick?

17. anonymous

can u calculate $$x_1 x_2$$? give it a try :-)

18. anonymous

One sec

19. anonymous

c/a

20. anonymous

you are great :-)

21. anonymous

But I still don't see how can I use that in this particular question :D

22. anonymous

$\frac{ x _{1} }{ x _{2} }=k$ That can be useful, should I try that?

23. anonymous

now you have three equations:$x_1+x_2=-\frac{b}{a}$$x_1x_2=\frac{c}{a}$$x_1=kx_2$play around 3 equations, tell me what you get? :-)

24. anonymous

I started with x1=kx2 For x1 I used x1=-b/a - x2 For x2 I used x2=(c/a)/x1 That way I used all three of them. Than I changed x1 and x2 with -b formula. Now I'm left with: $2b ^{2}-4ac-2b \sqrt{b ^{2}-4ac}=4kca$

25. anonymous

I don't see the way out

26. anonymous

you don't need to involve b formula again :-)

27. anonymous

I also tried x1/x2 with -b formula straight away but it get even more complicated.

28. anonymous

How? What do I do with x1 and x2

29. anonymous

I'll give you a hint, put $$x_1=kx_2$$ in first equation$x_1+x_2=-\frac{b}{a}$$kx_2+x_2=-\frac{b}{a}$$x_2(1+k)=-\frac{b}{a}$square both sides$x_2^2(1+k)^2=\frac{b^2}{a^2} \ \ \ \star$now calculate $$x_2^2$$ from 2 equations $$x_1=kx_2$$ and $$x_1x_2=\frac{c}{a}$$ and put it in the $$\star$$

30. anonymous

please show me the steps :-)

31. anonymous

This is ridiculous, now I got that $b ^{2}=c^{2}$ http://prntscr.com/3n5k5k

32. anonymous

try again :-) you are close to the answer

33. anonymous

check your steps again, be careful :-)

34. anonymous

No improvements

35. anonymous

is this equation clear for you: $x_2^2(1+k)^2=\frac{b^2}{a^2} \ \ \ \star$

36. anonymous

Seriously, I can't spot the mistake

37. anonymous

Absolutely

38. anonymous

ok, i'll solve it for you :-) $x_1=kx_2$$x_1x_2=\frac{c}{a}$put first in the second one$kx_2x_2=\frac{c}{a}$$x_2^2=\frac{c}{ka}$put this in the $$\star$$$x_2^2(1+k)^2=\frac{b^2}{a^2} \ \ \ \star$$\frac{c}{ka}(1+k)^2=\frac{b^2}{a^2} \ \ \ \star$$ac(1+k)^2=kb^2$

39. anonymous

let me know if there is a doubt on it :-)

40. anonymous

I see it now, I literally ignored multiplication. Thank you :) I got it now