If for some p , q , r ∈ R, not all have same sign, one of the roots of the equation (p2+q2) x2-2 q(p+r) x+q2+r2=0 is also a root of the equation x2+2 x-8=0, then ( q 2+ r 2/ p 2) is equal to-

Question

If for some p , q , r ∈ R, not all have same sign, one of the roots of the equation (p2+q2) x2-2 q(p+r) x+q2+r2=0 is also a root of the equation x2+2 x-8=0, then ( q 2+ r 2/ p 2) is equal to- (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

( px - q )2+( qx - r )2=0 ⇒ x =( q / p )=( r / q )=-4 ⇒ ( q 2+ r 2/ p 2)=272

Q. If for some $p, q, r \in R$ , not all have same 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 $\frac{q ^{2} + r ^{2}}{p ^{2}}$ is equal to-

$(p x - q)^{2} + (q x - r)^{2} = 0$
$\Rightarrow x = \frac{q}{p} = \frac{r}{q} = - 4$
$\Rightarrow \frac{q ^{2} + r ^{2}}{p ^{2}} = 272$

Q. If for some p,q,r∈R, not all have same sign, one of the roots of the equation (p2+q2)x2−2q(p+r)x+q2+r2=0 is also a root of the equation x2+2x−8=0, then p2q2+r2​ is equal to-

(px−q)2+(qx−r)2=0 ⇒x=pq​=qr​=−4 ⇒p2q2+r2​=272

Q. If for some $p, q, r \in R$ , not all have same 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 $\frac{q ^{2} + r ^{2}}{p ^{2}}$ is equal to-

$(p x - q)^{2} + (q x - r)^{2} = 0$
$\Rightarrow x = \frac{q}{p} = \frac{r}{q} = - 4$
$\Rightarrow \frac{q ^{2} + r ^{2}}{p ^{2}} = 272$