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nickersia Group Title

Show that kb^2=(k+1)^2*ac

  • 3 months ago
  • 3 months ago

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  1. mukushla Group Title
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    Hi :-) what is your constraint?

    • 3 months ago
  2. nickersia Group Title
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    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 months ago
  3. mukushla Group Title
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    have you tried it yet?

    • 3 months ago
  4. nickersia Group Title
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    \[\frac{ -b+\sqrt{b ^{2}-4ac} }{ 2a } = k \frac{ -b-\sqrt{b ^{2}-4ac} }{ 2a }\]

    • 3 months ago
  5. nickersia Group Title
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    That's how I started, but I got stuck, and it gets kinda messy

    • 3 months ago
  6. mukushla Group Title
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    let's play with variable roots, assume that roots are \(x_1\) and \(x_2\)

    • 3 months ago
  7. mukushla Group Title
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    what are the expressions for \(x_1 x_2\) and \(x_1+x_2\) in terms of quadratic coefficients?

    • 3 months ago
  8. nickersia Group Title
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    Not sure I understand what you mean? Two roots are in ratio 1:k, so x1=kx2

    • 3 months ago
  9. nickersia Group Title
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    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.

    • 3 months ago
  10. mukushla Group Title
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    do u know about formula for the Sum and product of the roots of a quadratic?

    • 3 months ago
  11. nickersia Group Title
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    Don't think so

    • 3 months ago
  12. mukushla Group Title
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    I will teach you then :-)

    • 3 months ago
  13. mukushla Group Title
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    well as you know\[x_1=\frac{ -b+\sqrt{b ^{2}-4ac} }{ 2a }\]\[x_2=\frac{ -b-\sqrt{b ^{2}-4ac} }{ 2a }\]

    • 3 months ago
  14. mukushla Group Title
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    now let's calculate \(x_1+x_2\): \[x_1+x_2=\frac{ -b+\sqrt{b ^{2}-4ac} }{ 2a }+\frac{ -b-\sqrt{b ^{2}-4ac} }{ 2a }\\=\frac{ -b+\sqrt{b ^{2}-4ac} -b-\sqrt{b ^{2}-4ac} }{ 2a }=-\frac{b}{a}\]

    • 3 months ago
  15. nickersia Group Title
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    Ok :) thats reasonably

    • 3 months ago
  16. mukushla Group Title
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    am i clear nick?

    • 3 months ago
  17. mukushla Group Title
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    can u calculate \(x_1 x_2\)? give it a try :-)

    • 3 months ago
  18. nickersia Group Title
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    One sec

    • 3 months ago
  19. nickersia Group Title
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    c/a

    • 3 months ago
  20. mukushla Group Title
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    you are great :-)

    • 3 months ago
  21. nickersia Group Title
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    But I still don't see how can I use that in this particular question :D

    • 3 months ago
  22. nickersia Group Title
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    \[\frac{ x _{1} }{ x _{2} }=k\] That can be useful, should I try that?

    • 3 months ago
  23. mukushla Group Title
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    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? :-)

    • 3 months ago
  24. nickersia Group Title
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    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: \[2b ^{2}-4ac-2b \sqrt{b ^{2}-4ac}=4kca\]

    • 3 months ago
  25. nickersia Group Title
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    I don't see the way out

    • 3 months ago
  26. mukushla Group Title
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    you don't need to involve b formula again :-)

    • 3 months ago
  27. nickersia Group Title
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    I also tried x1/x2 with -b formula straight away but it get even more complicated.

    • 3 months ago
  28. nickersia Group Title
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    How? What do I do with x1 and x2

    • 3 months ago
  29. mukushla Group Title
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    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 \(\star\)

    • 3 months ago
  30. mukushla Group Title
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    please show me the steps :-)

    • 3 months ago
  31. nickersia Group Title
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    This is ridiculous, now I got that \[b ^{2}=c^{2}\] http://prntscr.com/3n5k5k

    • 3 months ago
  32. mukushla Group Title
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    try again :-) you are close to the answer

    • 3 months ago
  33. mukushla Group Title
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    check your steps again, be careful :-)

    • 3 months ago
  34. nickersia Group Title
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    No improvements

    • 3 months ago
  35. mukushla Group Title
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    is this equation clear for you: \[x_2^2(1+k)^2=\frac{b^2}{a^2} \ \ \ \star\]

    • 3 months ago
  36. nickersia Group Title
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    Seriously, I can't spot the mistake

    • 3 months ago
  37. nickersia Group Title
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    Absolutely

    • 3 months ago
  38. mukushla Group Title
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    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 \(\star\)\[x_2^2(1+k)^2=\frac{b^2}{a^2} \ \ \ \star\]\[\frac{c}{ka}(1+k)^2=\frac{b^2}{a^2} \ \ \ \star\]\[ac(1+k)^2=kb^2\]

    • 3 months ago
  39. mukushla Group Title
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    let me know if there is a doubt on it :-)

    • 3 months ago
  40. nickersia Group Title
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    I see it now, I literally ignored multiplication. Thank you :) I got it now

    • 3 months ago
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