Question

find the condition that the ratio between the roots of the ax^2+bx+c=0 may me m:n

Answer 1

@satellite73

Answer 2

i don't understand the question

Answer 3

i am also

Answer 4

i min

Answer 5

it's simple really. there's a ratio. it's between roots. sprouting there innocently. conditionally, I may m is to n as...
get to work!

Answer 6

find the condition that the ratio between the roots of the equation ax^2+bx+c=0 may me m:n

Answer 7

didn't quite get that @wwe123 ...
could you repost it?

Answer 8

can u solve this more @Algebraic!

Answer 9

we know that if the roots are \(r_1\) and \(r_2\) then \(r_1r_2=\frac{c}{a}\) but somehow i don't think that answers the question
\(m\) and \(n\) could be anything.
are you suppossed to write \(m\) and \(n\) in terms of \(a,b,c\) or are you supposed to write \(a, b, c\) in terms of \(m,n\) ?

Answer 10

I certainly couldn't solve it any less, so I guess, yes, I can solve it more.

Answer 11

that is why they are called lessons

Answer 12

\[\frac{m}{n}=\frac{r_1}{r_2}\] you can find \(r_1,r_2\) using the quadratic formula if nothing else comes to mind

Answer 13

\[\frac{ m }{ n }=\frac{ -b+\sqrt{b ^{2}-4ac} }{-b-\sqrt{b ^{2}-4ac}}\]

Answer 14

@satellite73

Answer 15

or vice versa

Answer 16

Let the roots be my & ny
Therefore sum of roots ---> my+ny = -b/a
\((m+n)y = -b/a\) .......(1)
Product of roots---> my x ny = c/a
\(mny^2=c/a \).........(2)
Squaring (1), we get
\((m+n)^2 y^2 = (-b/a )^2 \)...........(3)
Dividing (3) by (2)…………(y^2 will be cancelled)
(m+n)^2/mn = b^2/a^2 x a/c
\(ac(m+n)^2 = b^2mn.\)
Hence, the required condition is \(ac(m+n)^2 = b^2mn.\)

Answer 17

thanks @hartnn

Answer 18

welcome ^_^
ask if u have any doubts in any step.

Answer 19

i use this method already
buy who u show that
Let the roots be my & ny
@hartnn

Answer 20

u mean *why?
since the roots are in ratio m:n, they are in ratio of any multiples of m and n,
like if the roots are in ratio 2:3
then they are in ratio 4:6 also

Answer 21

ok