Mathematics - Quadratic Equation Question with Solution | TestHub

We have ${\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3=-\mathrm{p}\dots$. (1)

$x_1{x}_2+x_1{x}_3+x_2{x}_3=q\dots \dots \left(2\right)$

$x_1{x}_2{x}_3=-r\dots \dots \left(3\right)$

$x_1^2=x_2{x}_3\dots \dots \left(4\right)$

from (2)

$$x_1{x}_2+x_1{x}_3+x_1^2=q$$

$$x_1\left(x_1+x_2+x_3\right)=q$$

$$x_1=-\frac{q}{p}$$

$$\begin{array}{rlrlr}& & & & \Rightarrow p{x}_1+q=0\Rightarrow K_1=1\end{array}$$

from (1)

$$\begin{array}{llc}{x}_2+x_3=\frac{q-p^2}{p}& \dots .x_2{x}_3=\frac{rp}{q}& \dots .\end{array}$$

Hence, ${\mathrm{x}}_2,{\mathrm{x}}_3$ satisfy the equation ${\mathrm{pqx}}^2-\left(\mathrm{q}-{\mathrm{p}}^2\right)\mathrm{qx}+{\mathrm{p}}^2\mathrm{r}=0$

$$\Rightarrow {\mathrm{K}}_2=1$$

From (5) and (8)

$${\mathrm{K}}_1+{\mathrm{K}}_2=2$$