Question

@ganeshie8 @BSwan

Answer 1

If one root of (l-m)x^2 +lx +1 =0 be double the other and if l be real , show that
\[m \le \frac{ 9 }{ 8 }\]

Answer 2

Do we have to substitute like let t = (l - m)

Answer 3

(l -m )a^2 = (l-m)4a^2 + al

Answer 4

Let the roots be \(\alpha, 2\alpha.\)\[(l-m)\alpha^2 + l\alpha + 1 = 0\]\[4(l-m)\alpha ^2 + 2l\alpha + 1 = 0\]\[\dfrac{\alpha^2}{l - 2l} = \dfrac{-\alpha}{(l-m) - 4(l-m)} = \dfrac{1}{(2l)(l-m) - (4l)(l-m)}\]That's the only thing which comes to my mind.

Answer 5

\( l\) is a natural log of what ?

Answer 6

there is no power given that's why i m confused

Answer 7

\[3\alpha = -\dfrac{l }{l - m}\]\[2\alpha^2 = \dfrac{1}{l - m}\]Plug the value of \(\alpha\) in the second equation...\[2\left(\dfrac{l}{3(l - m)}\right)^2 = \dfrac{1}{l - m}\]\[\dfrac{2l^2}{9(l - m)} = 1\]\[2l^2 = 9(l - m)\]\[\dfrac{l(9 - 2l)}{9} =m \]@ganeshie8 suspects that you haven't provided the full details in the question.

Answer 8

The only place where I can see the presence of inequalities is the discriminant.

But it's not given that the roots are real.

Answer 9

\(\alpha , 2\alpha\)

\(3\alpha = \dfrac{-l}{l-m}\)

\(2\alpha^2 = \dfrac{1}{l-m}\)

Answer 10

Yes that 's what @ParthKohli

Answer 11

yeah you can conclude by taking the discriminant >= 0

Answer 12

Why greater than or "="

Answer 13

But \(\alpha,2\alpha\) can be complex roots... no one knows. :|

Answer 14

cuz \(l\) is real, so the quadratic in \(l\) must have the discriminant >= 0

Answer 15

But if it's = 0, then it'd have two coincident roots. One root will be the double of the other only if the root is zero.

Answer 16

Yes there would be only one case possible

Answer 17

\(\alpha , 2\alpha\)

\(3\alpha = \dfrac{-l}{l-m} \implies 9\alpha^2 = \dfrac{l^2}{(l-m)^2}~~~ (1)\)
\(2\alpha^2 = \dfrac{1}{l-m} ~~~~(2)\)

\(\dfrac{(1)}{(2)} \implies \dfrac{9}{2} = \dfrac{l^2}{l-m} \implies 2l^2 - 9l + 9m=0
\)

Answer 18

Ah!

Answer 19

since \(l \in \mathbb R\), \(D \ge 0 \implies 9^2 - 4(2)(9m) \ge 0 \implies m \le \dfrac{9}{8} \)

Answer 20

Thanks! ur awesome thank u everybody

Answer 21

lol, didn't think of converting it to a quadratic.
I was close to that result.

Answer 22

that was beautiful :'(
ty @ganeshie8