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

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

mukushla Group TitleBest ResponseYou've already chosen the best response.2
have you tried it yet?
 6 months ago

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

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

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

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

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

nickersia Group TitleBest ResponseYou've already chosen the best response.0
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.
 6 months ago

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

nickersia Group TitleBest ResponseYou've already chosen the best response.0
Don't think so
 6 months ago

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

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

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

nickersia Group TitleBest ResponseYou've already chosen the best response.0
Ok :) thats reasonably
 6 months ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
am i clear nick?
 6 months ago

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

mukushla Group TitleBest ResponseYou've already chosen the best response.2
you are great :)
 6 months ago

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

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

mukushla Group TitleBest ResponseYou've already chosen the best response.2
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? :)
 6 months ago

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

nickersia Group TitleBest ResponseYou've already chosen the best response.0
I don't see the way out
 6 months ago

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

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

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

mukushla Group TitleBest ResponseYou've already chosen the best response.2
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 \(\star\)
 6 months ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
please show me the steps :)
 6 months ago

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

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

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

nickersia Group TitleBest ResponseYou've already chosen the best response.0
No improvements
 6 months ago

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

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

nickersia Group TitleBest ResponseYou've already chosen the best response.0
Absolutely
 6 months ago

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

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

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