JEE MAIN - Mathematics (2022 - 26th July Morning Shift - No. 15)
If for some $$\mathrm{p}, \mathrm{q}, \mathrm{r} \in \mathbf{R}$$, not all have same sign, one of the roots of the equation $$\left(\mathrm{p}^{2}+\mathrm{q}^{2}\right) x^{2}-2 \mathrm{q}(\mathrm{p}+\mathrm{r}) x+\mathrm{q}^{2}+\mathrm{r}^{2}=0$$ is also a root of the equation $$x^{2}+2 x-8=0$$, then $$\frac{\mathrm{q}^{2}+\mathrm{r}^{2}}{\mathrm{p}^{2}}$$ is equal to ____________,
Answer
272
Explanation
Let roots of
_26th_July_Morning_Shift_en_15_1.png)
$$\therefore$$ $$\alpha$$ + $$\beta$$ > 0 and $$\alpha$$$$\beta$$ > 0
Also, it has a common root with $${x^2} + 2x - 8 = 0$$
$$\therefore$$ The common root between above two equations is 4.
$$ \Rightarrow 16({p^2} + {q^2}) - 8q(p + r) + {q^2} + {r^2} = 0$$
$$ \Rightarrow (16{p^2} - 8pq + {q^2}) + (16{q^2} - 8qr + {r^2}) = 0$$
$$ \Rightarrow {(4p - q)^2} + {(4q - r)^2} = 0$$
$$ \Rightarrow q = 4p$$ and $$r = 16p$$
$$\therefore$$ $${{{q^2} + {r^2}} \over {{p^2}}} = {{16{p^2} + 256{p^2}} \over {{p^2}}} = 272$$
Comments (0)
