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Question

(D) Let the correct equation be $x^2 + px + q = 0$ $(i)$.The first student uses a wrong value of $p$ but the correct value of $q$. The roots found are

Vedclass · Accepted Answer

$-3, -4$

Vedclass · Answer

(D) Let the correct equation be $x^2 + px + q = 0$ $(i)$.The first student uses a wrong value of $p$ but the correct value of $q$. The roots found are $2$ and $6$.The product of roots is $2 	imes 6 = 12$. Since the constant term $q$ is correct,$q = 12$.The second student uses a wrong value of $q$ but the correct value of $p$. The roots found are $2$ and $-9$.The sum of roots is $2 + (-9) = -7$. Since the coefficient of $x$ $(p)$ is correct,$-p = -7$,which means $p = 7$.Substituting $p = 7$ and $q = 12$ into the original equation $(i)$,we get $x^2 + 7x + 12 = 0$.Factoring the quadratic equation: $x^2 + 4x + 3x + 12 = 0$ $\Rightarrow x(x + 4) + 3(x + 4) = 0$ $\Rightarrow (x + 3)(x + 4) = 0$.Thus,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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