Use Viette's formulae: " the product of both roots of a quadratic eq. equals its free coefficient divided by its leading coef.,
and the sum of the roots equals minus its linear coef. divided by its leading one".
Thus, $\displaystyle \displaystyle{\frac{m^2}{ln}=\frac{\left(l\lambda( \lambda+1)^2+(\lambda+1)^2\right)^2}{ln}=\frac{l(\ lambda+1)^6}{n}}$
$\displaystyle \displaystyle{\frac{b^2}{ac}=\frac{\left(-a(\alpha+\beta)\right)^2}{a^2\alpha\beta}=\frac{(\ alpha+\beta)^2}{\alpha\beta}=\frac{(\lambda\beta+\ beta)^2}{\lambda\beta^2}=\frac{(\lambda+1)^2}{\lam bda}}$.
Well, now just show that both expressions in the RHS above are equal (Viete again),
and for part (ii) is similar.
Tonio
$\displaystyle \displaystyle ax^2 + bx + c = 0$
Thus
$\displaystyle \displaystyle a \left ( x^2 + \frac{b}{a}x + \frac{c}{a} \right ) = 0$
If the two roots are $\displaystyle \beta$ and $\displaystyle \lambda \beta$ then we know that
$\displaystyle \displaystyle a ( x - \beta ) (x - \lambda \beta ) = 0$
Expanding this and equating powers of x from your a, b, c equation form gives
$\displaystyle \displaystyle \frac{b}{a} = - \beta ( \lambda + 1 )$
and
$\displaystyle \displaystyle \frac{c}{a} = \lambda \beta ^2$
Solve the first of these for b and square it:
$\displaystyle \displaystyle b^2 = a^2 \beta ( \lambda + 1 )^2$
So
$\displaystyle \displaystyle \frac{b^2}{a} = a \beta ( \lambda + 1 )^2$
and since
$\displaystyle \displaystyle c = a \lambda \beta ^2$
we have, after some simplifying:
$\displaystyle \displaystyle \frac{b^2}{ac} = \frac{( \lambda + 1 )^2}{\lambda}$
You do the l, m, n quadratic the same way.
-Dan
If α and β are the roots of ax^2 + bx + c = 0, then the sum of the roots = α + β = - b/a and product of the roots = αβ = c/a.
In the second equation,
(α^2 + β^2) + (α^2 - β^2) = -p and (α^2 + β^2) * (α^2 - β^2) = q.....(1)
Now (α^2 + β^2) = (α + β)^2 - 2αβ = ...?
(α - β)^2 = (α + β)^2 - 4αβ =.......? Find (α - β)
(α^2 - β^2) = (α + β)(α - β) = ......
Substitute these values in eq.(1)