Question

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

Answer 1

Hi :-)
what is your constraint?

Answer 2

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

Answer 3

have you tried it yet?

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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.

Answer 10

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

Answer 11

Don't think so

Answer 12

I will teach you then :-)

Answer 13

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

Answer 14

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

Answer 15

Ok :) thats reasonably

Answer 16

am i clear nick?

Answer 17

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

Answer 18

One sec

Answer 19

c/a

Answer 20

you are great :-)

Answer 21

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

Answer 22

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

Answer 23

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? :-)

Answer 24

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

Answer 25

I don't see the way out

Answer 26

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

Answer 27

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

Answer 28

How? What do I do with x1 and x2

Answer 29

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 \(\star\)

Answer 30

please show me the steps :-)

Answer 31

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

Answer 32

try again :-) you are close to the answer

Answer 33

check your steps again, be careful :-)

Answer 34

No improvements

Answer 35

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

Answer 36

Seriously, I can't spot the mistake

Answer 37

Absolutely

Answer 38

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

Answer 39

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

Answer 40

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