A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 2 years ago
Show that kb^2=(k+1)^2*ac
anonymous
 2 years ago
Show that kb^2=(k+1)^2*ac

This Question is Closed

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Hi :) what is your constraint?

anonymous
 2 years 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 \[kb ^{2}=(k+1)^{2}ac\]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0have you tried it yet?

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

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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0let's play with variable roots, assume that roots are \(x_1\) and \(x_2\)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0what are the expressions for \(x_1 x_2\) and \(x_1+x_2\) in terms of quadratic coefficients?

anonymous
 2 years 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

anonymous
 2 years 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.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0do u know about formula for the Sum and product of the roots of a quadratic?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I will teach you then :)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0well as you know\[x_1=\frac{ b+\sqrt{b ^{2}4ac} }{ 2a }\]\[x_2=\frac{ b\sqrt{b ^{2}4ac} }{ 2a }\]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0now 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
 2 years ago
Best ResponseYou've already chosen the best response.0Ok :) thats reasonably

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0can u calculate \(x_1 x_2\)? give it a try :)

anonymous
 2 years 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

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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0now 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
 2 years 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: \[2b ^{2}4ac2b \sqrt{b ^{2}4ac}=4kca\]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I don't see the way out

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0you don't need to involve b formula again :)

anonymous
 2 years 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.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0How? What do I do with x1 and x2

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I'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
 2 years ago
Best ResponseYou've already chosen the best response.0please show me the steps :)

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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0try again :) you are close to the answer

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0check your steps again, be careful :)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0is this equation clear for you: \[x_2^2(1+k)^2=\frac{b^2}{a^2} \ \ \ \star\]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Seriously, I can't spot the mistake

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0ok, 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
 2 years ago
Best ResponseYou've already chosen the best response.0let me know if there is a doubt on it :)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I see it now, I literally ignored multiplication. Thank you :) I got it now
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.