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

- anonymous

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

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- anonymous

Hi :-)
what is your constraint?

- anonymous

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

- anonymous

have you tried it yet?

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## More answers

- anonymous

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

- anonymous

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

- anonymous

let's play with variable roots, assume that roots are \(x_1\) and \(x_2\)

- anonymous

what are the expressions for \(x_1 x_2\) and \(x_1+x_2\) in terms of quadratic coefficients?

- anonymous

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

- 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.

- anonymous

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

- anonymous

Don't think so

- anonymous

I will teach you then :-)

- anonymous

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

- anonymous

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

- anonymous

Ok :) thats reasonably

- anonymous

am i clear nick?

- anonymous

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

- anonymous

One sec

- anonymous

c/a

- anonymous

you are great :-)

- anonymous

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

- anonymous

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

- 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? :-)

- 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:
\[2b ^{2}-4ac-2b \sqrt{b ^{2}-4ac}=4kca\]

- anonymous

I don't see the way out

- anonymous

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

- anonymous

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

- anonymous

How? What do I do with x1 and x2

- 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 \(\star\)

- anonymous

please show me the steps :-)

- anonymous

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

- anonymous

try again :-) you are close to the answer

- anonymous

check your steps again, be careful :-)

- anonymous

No improvements

- anonymous

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

- anonymous

Seriously, I can't spot the mistake

- anonymous

Absolutely

- 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 \(\star\)\[x_2^2(1+k)^2=\frac{b^2}{a^2} \ \ \ \star\]\[\frac{c}{ka}(1+k)^2=\frac{b^2}{a^2} \ \ \ \star\]\[ac(1+k)^2=kb^2\]

- anonymous

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

- anonymous

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

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