If the roots of the equation a²+bx+c = 0 are of the form$\frac{\mathrm{k}+1}{\mathrm{k}} \mathrm{and} \frac{\mathrm{k}+2}{\mathrm{k}+1}$, then ${\left(a+b+c\right)}^2$ is equal to

Solution:

we have.$\frac{\mathrm{k}+1}{\mathrm{k}}+\frac{\mathrm{k}+2}{\mathrm{k}+1}=-\frac{\mathrm{b}}{\mathrm{a}}..................\left(\mathrm{i}\right)$
and $\frac{\mathrm{k}+1}{\mathrm{k}}\cdot \frac{\mathrm{k}+2}{\mathrm{k}+1}=\frac{\mathrm{c}}{\mathrm{a}}$……………………….(ii)
From Eq. (i), 
$1+\frac{1}{\mathrm{k}}+1+\frac{1}{\mathrm{k}+1}=-\frac{\mathrm{b}}{\mathrm{a}}$
$\Rightarrow 2+\frac{1}{\mathrm{k}}+\frac{1}{\mathrm{k}+1}=-\frac{\mathrm{b}}{\mathrm{a}}$…………………..(iii)
From Eq. (ii),
$\begin{array}{r}\frac{\mathrm{k}+2}{\mathrm{k}}=\frac{\mathrm{c}}{\mathrm{a}}\Rightarrow 1+\frac{2}{\mathrm{k}}=\frac{\mathrm{c}}{\mathrm{a}}\\ \Rightarrow \frac{2}{\mathrm{k}}=\frac{\mathrm{c}}{\mathrm{a}}-1\Rightarrow \mathrm{k}=\frac{2\mathrm{a}}{\mathrm{c}-\mathrm{a}}\end{array}$
Now, on substituting the value of kin Eq. (iii), we get
$2+\frac{\mathrm{c}-\mathrm{a}}{2\mathrm{a}}+\frac{1}{\frac{2\mathrm{a}}{\mathrm{c}-\mathrm{a}}+1}=-\frac{\mathrm{b}}{\mathrm{a}}$

$\Rightarrow 2+\frac{\mathrm{c}-\mathrm{a}}{2\mathrm{a}}+\frac{\mathrm{c}-\mathrm{a}}{\mathrm{a}+\mathrm{c}}=-\frac{b}{a}$
$\begin{array}{l}\Rightarrow \frac{2\left(2\mathrm{a}\right)(\mathrm{a}+\mathrm{c})+(\mathrm{c}-\mathrm{a})(\mathrm{c}+\mathrm{a})+2\mathrm{a}(\mathrm{c}-\mathrm{a})}{2\mathrm{a}(\mathrm{a}+\mathrm{c})}=-\frac{\mathrm{b}}{\mathrm{a}}\\ \Rightarrow {\mathrm{a}}^2+{\mathrm{c}}^2+6\mathrm{ac}=-2\mathrm{ab}-2\mathrm{bc}\\ \Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2+2\mathrm{ab}+2\mathrm{bc}+2\mathrm{ca}={\mathrm{b}}^2-4\mathrm{ac}\\ \therefore (\mathrm{a}+\mathrm{b}+\mathrm{c}{)}^2={\mathrm{b}}^2-4\mathrm{ac}\end{array}$