If for some p, q, r in mathbb{R},not all have | vedclass

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(D) The given equation is $(p^{2}+q^{2})x^{2}-2q(p+r)x + q^{2}+r^{2}=0$.This can be rewritten as $p^{2}x^{2} - 2pqx + q^{2} + q^{2}x^{2} - 2qrx + r^{2

Vedclass · Accepted Answer

$272$

Vedclass · Answer

(D) The given equation is $(p^{2}+q^{2})x^{2}-2q(p+r)x + q^{2}+r^{2}=0$.This can be rewritten as $p^{2}x^{2} - 2pqx + q^{2} + q^{2}x^{2} - 2qrx + r^{2} = 0$.This simplifies to $(px-q)^{2} + (qx-r)^{2} = 0$.Since $p, q, r \in \mathbb{R}$,the sum of squares is zero only if each term is zero:$px-q = 0 \implies x = \frac{q}{p}$ and $qx-r = 0 \implies x = \frac{r}{q}$.Thus,the root is $x = \frac{q}{p} = \frac{r}{q}$.The second equation is $x^{2}+2x-8=0$,which factors as $(x+4)(x-2)=0$,so $x = -4$ or $x = 2$.If $x = 2$,then $\frac{q}{p} = 2 \implies q = 2p$ and $\frac{r}{q} = 2 \implies r = 2q = 4p$.Then $\frac{q^{2}+r^{2}}{p^{2}} = \frac{(2p)^{2} + (4p)^{2}}{p^{2}} = \frac{4p^{2} + 16p^{2}}{p^{2}} = 20$.If $x = -4$,then $\frac{q}{p} = -4 \implies q = -4p$ and $\frac{r}{q} = -4 \implies r = -4q = 16p$.Then $\frac{q^{2}+r^{2}}{p^{2}} = \frac{(-4p)^{2} + (16p)^{2}}{p^{2}} = \frac{16p^{2} + 256p^{2}}{p^{2}} = 272$.Since the problem states $p, q, r$ do not all have the same sign,for $x=-4$,$q=-4p$ and $r=16p$ satisfy this condition (e.g.,$p=1, q=-4, r=16$).Thus,the value is $272$.

If for some $p, q, r \in \mathbb{R}$,not all have the same sign,and one of the roots of the equation $(p^{2}+q^{2})x^{2}-2q(p+r)x + q^{2}+r^{2}=0$ is also a root of the equation $x^{2}+2x-8=0$,then $\frac{q^{2}+r^{2}}{p^{2}}$ is equal to-

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