anonymous
  • anonymous
Show that kb^2=(k+1)^2*ac
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Hi :-) what is your constraint?
anonymous
  • 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
  • anonymous
have you tried it yet?

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anonymous
  • anonymous
\[\frac{ -b+\sqrt{b ^{2}-4ac} }{ 2a } = k \frac{ -b-\sqrt{b ^{2}-4ac} }{ 2a }\]
anonymous
  • anonymous
That's how I started, but I got stuck, and it gets kinda messy
anonymous
  • anonymous
let's play with variable roots, assume that roots are \(x_1\) and \(x_2\)
anonymous
  • anonymous
what are the expressions for \(x_1 x_2\) and \(x_1+x_2\) in terms of quadratic coefficients?
anonymous
  • anonymous
Not sure I understand what you mean? Two roots are in ratio 1:k, so x1=kx2
anonymous
  • 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
  • anonymous
do u know about formula for the Sum and product of the roots of a quadratic?
anonymous
  • anonymous
Don't think so
anonymous
  • anonymous
I will teach you then :-)
anonymous
  • anonymous
well as you know\[x_1=\frac{ -b+\sqrt{b ^{2}-4ac} }{ 2a }\]\[x_2=\frac{ -b-\sqrt{b ^{2}-4ac} }{ 2a }\]
anonymous
  • 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
  • anonymous
Ok :) thats reasonably
anonymous
  • anonymous
am i clear nick?
anonymous
  • anonymous
can u calculate \(x_1 x_2\)? give it a try :-)
anonymous
  • anonymous
One sec
anonymous
  • anonymous
c/a
anonymous
  • anonymous
you are great :-)
anonymous
  • anonymous
But I still don't see how can I use that in this particular question :D
anonymous
  • anonymous
\[\frac{ x _{1} }{ x _{2} }=k\] That can be useful, should I try that?
anonymous
  • 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
  • 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
  • anonymous
I don't see the way out
anonymous
  • anonymous
you don't need to involve b formula again :-)
anonymous
  • anonymous
I also tried x1/x2 with -b formula straight away but it get even more complicated.
anonymous
  • anonymous
How? What do I do with x1 and x2
anonymous
  • 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
  • anonymous
please show me the steps :-)
anonymous
  • anonymous
This is ridiculous, now I got that \[b ^{2}=c^{2}\] http://prntscr.com/3n5k5k
anonymous
  • anonymous
try again :-) you are close to the answer
anonymous
  • anonymous
check your steps again, be careful :-)
anonymous
  • anonymous
No improvements
anonymous
  • anonymous
is this equation clear for you: \[x_2^2(1+k)^2=\frac{b^2}{a^2} \ \ \ \star\]
anonymous
  • anonymous
Seriously, I can't spot the mistake
anonymous
  • anonymous
Absolutely
anonymous
  • 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
  • anonymous
let me know if there is a doubt on it :-)
anonymous
  • anonymous
I see it now, I literally ignored multiplication. Thank you :) I got it now

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