If the equations x^2+px+2=0 and x^2+x+2p=0 ha | vedclass

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(C) Let $\alpha$ be the common root of the equations $x^2+px+2=0$ and $x^2+x+2p=0$. Then,$\alpha^2+p\alpha+2=0$ and $\alpha^2+\alpha+2p=0$. Subtractin

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$6$

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(C) Let $\alpha$ be the common root of the equations $x^2+px+2=0$ and $x^2+x+2p=0$. Then,$\alpha^2+p\alpha+2=0$ and $\alpha^2+\alpha+2p=0$. Subtracting the two equations,we get $(p-1)\alpha + (2-2p) = 0$,which simplifies to $(p-1)\alpha - 2(p-1) = 0$. This implies $(p-1)(\alpha-2) = 0$. Case $1$: If $p=1$,the equations become $x^2+x+2=0$ and $x^2+x+2=0$,which are identical. However,the discriminant $D = 1^2 - 4(1)(2) = -7 Case $2$: If $\alpha=2$,substituting $\alpha=2$ into $x^2+px+2=0$ gives $4+2p+2=0$,so $2p = -6$,which means $p=-3$. Now,substitute $p=-3$ into the equation $x^2+2px+8=0$ to get $x^2+2(-3)x+8=0$,which is $x^2-6x+8=0$. The sum of the roots of a quadratic equation $ax^2+bx+c=0$ is given by $-b/a$. For $x^2-6x+8=0$,the sum of the roots is $-(-6)/1 = 6$.

If the equations $x^2+px+2=0$ and $x^2+x+2p=0$ have a common root,then the sum of the roots of the equation $x^2+2px+8=0$ is

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