If for some p , q , r ∈ R , all have positive sign, one of the roots of the equation p 2 + q 2 x 2 - 2 q p + r x + q 2 + r 2 = 0 is also a root of the equation x 2 + 2 x - 8 = 0 , then q 2 + r 2 p 2 is equal to-

Question

If for some p , q , r &#8712; R , all have positive sign, one of the roots of the equation p 2 + q 2 x 2 - 2 q p + r x + q 2 + r 2 = 0 is also a root of the equation x 2 + 2 x - 8 = 0 , then q 2 + r 2 p 2 is equal to-

Aicharya · Accepted Answer

Given, $(p^2 + q^2)x^2 - 2q(p + r)x + q^2 + r^2 = 0$ On simplifying we get, $(px - q)^2 + (qx - r)^2 = 0$ $\Rightarrow px - q = 0 \quad \& \quad qx - r = 0$ $\Rightarrow x = \frac{q}{p} = \frac{r}{q}$ $\Rightarrow x = \frac{q}{p} = \frac{r}{q} = 4$ [because roots of equation $x^2 - 2x - 8 = 0$ are $4$ or $-2$] As $p, q, r$ are positive, so $x$ must be $4$. Now, $q = 4p$ and $r = 4q = 16p$ So, $\frac{q^2 + r^2}{p^2} = \frac{(4p)^2 + (16p)^2}{p^2}$ = 16 + 256 = 272

If for some p , q , r ∈ R , all have positive sign, one of the roots of the equation p 2 + q 2 x 2 - 2…

Solution

If for some p , q , r &#8712; R , all have positive sign, one of the roots of the equation p 2 + q 2 x 2 - 2…

Solution

If for some p , q , r ∈ R , all have positive sign, one of the roots of the equation p 2 + q 2 x 2 - 2…