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nickersia
 10 months ago
Show that kb^2=(k+1)^2*ac
nickersia
 10 months ago
Show that kb^2=(k+1)^2*ac

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mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2Hi :) what is your constraint?

nickersia
 10 months ago
Best ResponseYou've already chosen the best response.0One of the roots of the quadratic equation \[ax ^{2}+bx+c=0\] is k times the other root. Show that \[kb ^{2}=(k+1)^{2}ac\]

mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2have you tried it yet?

nickersia
 10 months ago
Best ResponseYou've already chosen the best response.0\[\frac{ b+\sqrt{b ^{2}4ac} }{ 2a } = k \frac{ b\sqrt{b ^{2}4ac} }{ 2a }\]

nickersia
 10 months ago
Best ResponseYou've already chosen the best response.0That's how I started, but I got stuck, and it gets kinda messy

mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2let's play with variable roots, assume that roots are \(x_1\) and \(x_2\)

mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2what are the expressions for \(x_1 x_2\) and \(x_1+x_2\) in terms of quadratic coefficients?

nickersia
 10 months ago
Best ResponseYou've already chosen the best response.0Not sure I understand what you mean? Two roots are in ratio 1:k, so x1=kx2

nickersia
 10 months ago
Best ResponseYou've already chosen the best response.0I 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.

mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2do u know about formula for the Sum and product of the roots of a quadratic?

mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2I will teach you then :)

mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2well as you know\[x_1=\frac{ b+\sqrt{b ^{2}4ac} }{ 2a }\]\[x_2=\frac{ b\sqrt{b ^{2}4ac} }{ 2a }\]

mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2now let's calculate \(x_1+x_2\): \[x_1+x_2=\frac{ b+\sqrt{b ^{2}4ac} }{ 2a }+\frac{ b\sqrt{b ^{2}4ac} }{ 2a }\\=\frac{ b+\sqrt{b ^{2}4ac} b\sqrt{b ^{2}4ac} }{ 2a }=\frac{b}{a}\]

nickersia
 10 months ago
Best ResponseYou've already chosen the best response.0Ok :) thats reasonably

mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2can u calculate \(x_1 x_2\)? give it a try :)

nickersia
 10 months ago
Best ResponseYou've already chosen the best response.0But I still don't see how can I use that in this particular question :D

nickersia
 10 months ago
Best ResponseYou've already chosen the best response.0\[\frac{ x _{1} }{ x _{2} }=k\] That can be useful, should I try that?

mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2now 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? :)

nickersia
 10 months ago
Best ResponseYou've already chosen the best response.0I 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: \[2b ^{2}4ac2b \sqrt{b ^{2}4ac}=4kca\]

nickersia
 10 months ago
Best ResponseYou've already chosen the best response.0I don't see the way out

mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2you don't need to involve b formula again :)

nickersia
 10 months ago
Best ResponseYou've already chosen the best response.0I also tried x1/x2 with b formula straight away but it get even more complicated.

nickersia
 10 months ago
Best ResponseYou've already chosen the best response.0How? What do I do with x1 and x2

mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2I'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 \(\star\)

mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2please show me the steps :)

nickersia
 10 months ago
Best ResponseYou've already chosen the best response.0This is ridiculous, now I got that \[b ^{2}=c^{2}\] http://prntscr.com/3n5k5k

mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2try again :) you are close to the answer

mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2check your steps again, be careful :)

mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2is this equation clear for you: \[x_2^2(1+k)^2=\frac{b^2}{a^2} \ \ \ \star\]

nickersia
 10 months ago
Best ResponseYou've already chosen the best response.0Seriously, I can't spot the mistake

mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2ok, 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 \(\star\)\[x_2^2(1+k)^2=\frac{b^2}{a^2} \ \ \ \star\]\[\frac{c}{ka}(1+k)^2=\frac{b^2}{a^2} \ \ \ \star\]\[ac(1+k)^2=kb^2\]

mukushla
 10 months ago
Best ResponseYou've already chosen the best response.2let me know if there is a doubt on it :)

nickersia
 10 months ago
Best ResponseYou've already chosen the best response.0I see it now, I literally ignored multiplication. Thank you :) I got it now
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