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(D) Let the correct equation be $x^2 + px + q = 0$ .....$(i)$For the first candidate,the roots are $2$ and $6$. Since the error is only in $p$,the con

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$-3, -4$

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(D) Let the correct equation be $x^2 + px + q = 0$ .....$(i)$For the first candidate,the roots are $2$ and $6$. Since the error is only in $p$,the constant term $q$ is correct.Product of roots $q = 2 	imes 6 = 12$.For the second candidate,the roots are $2$ and $-9$. Since the error is only in $q$,the coefficient $p$ is correct.Sum of roots $= 2 + (-9) = -7 = -p$,which implies $p = 7$.Substituting the correct values of $p$ and $q$ into equation $(i)$,we get:$x^2 + 7x + 12 = 0$Factorizing the quadratic equation:$x^2 + 4x + 3x + 12 = 0$$x(x + 4) + 3(x + 4) = 0$$(x + 3)(x + 4) = 0$Therefore,the roots are $x = -3$ and $x = -4$.

Two candidates attempt to solve the equation $x^2 + px + q = 0$. One starts with a wrong value of $p$ and finds the roots to be $2$ and $6$,and the other starts with a wrong value of $q$ and finds the roots to be $2$ and $-9$. The roots of the original equation are

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