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 \fracq^2+r^2p^2 is equal to \underline\hspace2cm ,

Quadratic Equation and Inequalities PYQ — JEE Main 2022 · 26 Jul · Shift 1 · Q39 with Solution

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Correct answer: 272

Given $x^2+2x-8=0$ we factorize: $x^2+2x-8=(x-2)(x+4)=0$ So its roots are

\quad x=-4.$$ We are told that **one of these** is also a root of $$(p^2+q^2)x^2-2q(p+r)x+(q^2+r^2)=0.$$ Let $$A=p^2+q^2,\quad B=-2q(p+r),\quad C=q^2+r^2.$$ Then the equation is $$Ax^2+Bx+C=0.$$ --- 2. **Use the condition that one root is 2 or -4** Substitute each possibility. ### Case 1: $x=2$ Then $$4(p^2+q^2)-4q(p+r)+(q^2+r^2)=0.$$ Simplify: $$4p^2+4q^2-4pq-4qr+q^2+r^2=0$$ $$4p^2-4pq+5q^2-4qr+r^2=0.$$ Now observe that $$4p^2-4pq+q^2=(2p-q)^2$$ and $$4q^2-4qr+r^2=(2q-r)^2.$$ So $$4p^2-4pq+5q^2-4qr+r^2=(2p-q)^2+(2q-r)^2=0.$$ Hence both squares must be zero: $$2p-q=0,\quad 2q-r=0.$$ Thus $$q=2p,\quad r=2q=4p.$$ So $p,q,r$ all have the same sign (or all zero if $p=0$), which contradicts the condition **not all have same sign**. Therefore, $x=2$ is **not possible**. --- ### Case 2: $x=-4$ Then $$16(p^2+q^2)+8q(p+r)+(q^2+r^2)=0.$$ Simplify: $$16p^2+16q^2+8pq+8qr+q^2+r^2=0$$ $$16p^2+8pq+17q^2+8qr+r^2=0.$$ Rewrite as $$16p^2+8pq+q^2+16q^2+8qr+r^2=0$$ $$=(4p+q)^2+(4q+r)^2=0.$$ Hence $$4p+q=0,\quad 4q+r=0.$$ Thus $$q=-4p,\quad r=-4q=16p.$$ Now compute $$\frac{q^2+r^2}{p^2}= rac{(-4p)^2+(16p)^2}{p^2}= rac{16p^2+256p^2}{p^2}=272.$$ --- 3. **Check sign condition** For $p\neq 0$, $$q=-4p,\quad r=16p,$$ so the signs are $p$, opposite sign for $q$, and same sign as $p$ for $r$. Hence they are indeed **not all of the same sign**. Thus this case is valid. --- 4. **Final answer** $$\boxed{272}$$

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Correct answer: 272

Given $x^2+2x-8=0$ we factorize: $x^2+2x-8=(x-2)(x+4)=0$ So its roots are

\quad x=-4.$$ We are told that **one of these** is also a root of $$(p^2+q^2)x^2-2q(p+r)x+(q^2+r^2)=0.$$ Let $$A=p^2+q^2,\quad B=-2q(p+r),\quad C=q^2+r^2.$$ Then the equation is $$Ax^2+Bx+C=0.$$ --- 2. **Use the condition that one root is 2 or -4** Substitute each possibility. ### Case 1: $x=2$ Then $$4(p^2+q^2)-4q(p+r)+(q^2+r^2)=0.$$ Simplify: $$4p^2+4q^2-4pq-4qr+q^2+r^2=0$$ $$4p^2-4pq+5q^2-4qr+r^2=0.$$ Now observe that $$4p^2-4pq+q^2=(2p-q)^2$$ and $$4q^2-4qr+r^2=(2q-r)^2.$$ So $$4p^2-4pq+5q^2-4qr+r^2=(2p-q)^2+(2q-r)^2=0.$$ Hence both squares must be zero: $$2p-q=0,\quad 2q-r=0.$$ Thus $$q=2p,\quad r=2q=4p.$$ So $p,q,r$ all have the same sign (or all zero if $p=0$), which contradicts the condition **not all have same sign**. Therefore, $x=2$ is **not possible**. --- ### Case 2: $x=-4$ Then $$16(p^2+q^2)+8q(p+r)+(q^2+r^2)=0.$$ Simplify: $$16p^2+16q^2+8pq+8qr+q^2+r^2=0$$ $$16p^2+8pq+17q^2+8qr+r^2=0.$$ Rewrite as $$16p^2+8pq+q^2+16q^2+8qr+r^2=0$$ $$=(4p+q)^2+(4q+r)^2=0.$$ Hence $$4p+q=0,\quad 4q+r=0.$$ Thus $$q=-4p,\quad r=-4q=16p.$$ Now compute $$\frac{q^2+r^2}{p^2}= rac{(-4p)^2+(16p)^2}{p^2}= rac{16p^2+256p^2}{p^2}=272.$$ --- 3. **Check sign condition** For $p\neq 0$, $$q=-4p,\quad r=16p,$$ so the signs are $p$, opposite sign for $q$, and same sign as $p$ for $r$. Hence they are indeed **not all of the same sign**. Thus this case is valid. --- 4. **Final answer** $$\boxed{272}$$