Question

if x1 and x2 are the roots of quadratic equation x^2+3mx+2n=0, then

Answer 1

question continues here
\[\left( \frac{ 1 }{ x _{1^{2}}} - \frac{ 1 }{ x _{2^{2}} } \right)^{2}\]

Answer 2

Are you supposed to find an expression for that in terms of \(m\) and \(n\)?

Answer 3

\[x^2+3mx+2n=0\]
has roots
\[x_{1,2}=\frac{-3m\pm\sqrt{9m^2-8n}}{2}\]
This gives you
\[\small \begin{align*}\left(\frac{1}{{x_1}^2}-\frac{1}{{x_2}^2}\right)^2&{\large=}\left(\left(\frac{2}{-3m+\sqrt{9m^2-8n}}\right)^2-\left(\frac{2}{-3m-\sqrt{9m^2-8n}}\right)^2\right)^2\\

&{\large=}\left(\frac{4}{9m^2-6m\sqrt{9m^2-8n}+9m^2-8n}-\frac{4}{9m^2+6m\sqrt{9m^2-8n}+9m^2-8n}\right)^2\\

&{\large=}\left(\frac{4(9m^2+6m\sqrt{9m^2-8n}+9m^2-8n)-4(9m^2-6m\sqrt{9m^2-8n}+9m^2-8n)}{(9m^2-6m\sqrt{9m^2-8n}+9m^2-8n)(9m^2+6m\sqrt{9m^2-8n}+9m^2-8n)}\right)^2\\

&{\large=}\left(\frac{48m\sqrt{9m^2-8n}}{64 n^2}\right)^2\\
&{\large=}\frac{2304m^2(9m^2-8n)}{4096 n^4}
\end{align*}\]

Answer 4

but the answers given are
A. \[\left( \frac{ 3m }{ 4n ^{2} } \right)^{2} \left( 9m ^{2}-8n \right)\]
B. \[\left( \frac{ 3m }{ 4n } \right)^{2} \left( 9m ^{2} -8n\right)\]
C. \[\left( \frac{ 3m }{ 4n } \right)^{2} \left( 9m ^{2}-8n ^{2} \right)\]
D. \[\left( \frac{ 3m ^{2} }{ 4n ^{2} } \right) \left( 9m ^{2}-8n ^{2} \right)\]
E. \[\left( \frac{ 3m }{ 4n } \right)^{2} \left( 9m ^{2}-8n \right)\]

Answer 5

See:

for an equation ax^2 + bx + c = 0 with roots u and v:

\[\large{u + v = \cfrac{-b}{a}}\]
\[\large{uv = \cfrac{c}{a}}\]

Answer 6

Comparing this equation with equation in question;

\[\large{x_1 + x_2 = -3m}\]

\[\large{x_1 * x_2 = 2n}\]

Answer 7

Now :

\[\large{(\cfrac{1}{{x_1}^2} - \cfrac{1}{{x_2}^2})^2}\]

\[\large{= \cfrac{{x_2}^2 - {x_1}^2}{{x_1}^2{x_2}^2}}\]

\[\large{= \cfrac{(x_2-x_1)(x_1+x_2)}{(x_1x_2)^2}}\]

\[\large{= \cfrac{(x_2-x_1)(-3m)}{(2n)^2}}\]

\[\large{= \cfrac{(x_1-x_2)(3m)}{4n^2}}\]

Answer 8

Now,

\[\large{(x_1-x_2)^2 = (x_1+x_2)^2 - 4x_1x_2}\]

\[\large{\implies (x_1-x_2)^2 = (-3m)^2 - 4(2n)}\]

\[\large{\implies (x_1-x_2)^2 = 9m^2 - 8n}\]

\[\large{\implies (x_1-x_2) = \sqrt{9m^2 - 8n}}\]

Answer 9

*** Please note: I am calculating only \(\large{(\cfrac{1}{x_1^2}-\cfrac{1}{x_2^2})}\). I by mistake added a complete square in the very first step.

Answer 10

Thus,

\[\large{(\cfrac{1}{x_1^2} - \cfrac{1}{x_2^2}) = \cfrac{\sqrt{9m^2-8n}(3m)}{4n^2}}\]

Thus,

\[\large{(\cfrac{1}{x_1^2}-\cfrac{1}{x_2^2})^2 = \cfrac{9m^2}{16n^4}(9m^2-8n)}\]

Answer 11

\[\large{implies (\cfrac{1}{x_1^2} - \cfrac{1}{x_2^2})^2 = (\cfrac{3m}{4n^2})^2(9m^2-8n)}\]

Answer 12

Thus, A is the correct option. :)

Answer 13

Did you get it @dinisha ?

Answer 14

perfect!! :) thank u @vishweshshrimali5

Answer 15

:) np