Question

If x1 and x2 are the roots of quadratic equation x^2+3mx+2n=0,then: ( 1/x1^2- 1/x2^2)=............

Answer 1

Help plz

Answer 2

Help plz

Answer 3

If \(x_1\) and \(x_2\) are roots of a quadratic, that means they are solutions to the equation \(x^2+3mx+2n =0\)

Now, you can factor a quadratic equation as \((x-a)(x-b)=0\) where \(a, b\) are the roots. So \((x-x_1)(x-x_2)=0\). Expand this:
\(x^2-xx_2-x_1x+x_1x_2=x^2-(x_1+x_2)x+x_1x_2\) This is equal to to the expression above.

Hence:
\( x^2-(x_1+x_2)x+x_1x_2=x^2+3mx+2n\) Just be careful that here, the variable is \(x\) are \(x_1, x_2\) are just roots, so they are numbers. By comparing terms, we see that:

\(3m=-(x_1+x_2)\)
\(2n=x_1x_2\)

Now solve for \(x_1\) and \(x_2\) in terms of \(m\) and \(n\) only:

\(3m = -x_1-x_2\)
\(-x_1=3m+x_2\)
\(x_1=-3m-x_2\)

\(x_2=-3m-x_1\)

and
\[ x_1=\frac{2n}{x_2}\\
x_2 = \frac{2n}{x_1}\]

Now plug in these results into the former equations.
\[ \frac{2n}{x_2}=-3m-x_2\\
2n=-3mx_2-x_2^2\\
x_2^2+3mx_2+2n=0\\
x_2=\frac{-(3m)\pm\sqrt{(3m)^2-4(1)(2n)}}{2(1)}=\frac{-3m \pm \sqrt{9m^2-8n}}{2}\]

The same for \(x_1\):
\[ \frac{2n}{x_1}=-3m-x_1\\
\\ \,\\

...\\
x_1=\frac{-(3m)\pm\sqrt{(3m)^2-4(1)(2n)}}{2(1)}=\frac{-3m \pm \sqrt{9m^2-8n}}{2}\]

So indeed we have that both roots give the same answer, so overall the 2 roots are \[ \frac{-3m \pm \sqrt{9m^2-4n}}{2}\]

Answer 4

Oops change that \(4n\) to \(8n\)

Answer 5

However it is not clear in your problem which root should be designated as \(x_1\) or \(x_2\). Presumably, \(x_1\) is the "first" root that appears in your parabola, so presumably you will find:
\[ \Large \frac{1}{\left(\frac{-3m-\sqrt{9m^2-8n}}{2}\right)^2} - \frac{1}{\left(\frac{-3m+\sqrt{9m^2-8n}}{2}\right)^2}\]