Two students were solving a quadratic equatio | vedclass

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(D) Let the correct quadratic equation be $x^2 + bx + c = 0$ (since the coefficient of $x^2$ is $1$).For the first student,the constant term was copie

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$6, -1$

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(D) Let the correct quadratic equation be $x^2 + bx + c = 0$ (since the coefficient of $x^2$ is $1$).For the first student,the constant term was copied incorrectly,but the coefficient of $x$ $(b)$ was copied correctly. The roots obtained were $3$ and $2$. Thus,the sum of roots is $-b = 3 + 2 = 5$,which implies $b = -5$.For the second student,the constant term $(c)$ and the coefficient of $x^2$ $(a=1)$ were copied correctly. Given $c = -6$ and $a = 1$,the product of the roots is $c/a = -6/1 = -6$.The correct quadratic equation is $x^2 - 5x - 6 = 0$.Factoring the equation: $x^2 - 6x + x - 6 = 0 \implies x(x - 6) + 1(x - 6) = 0 \implies (x - 6)(x + 1) = 0$.Therefore,the correct roots are $6$ and $-1$.

Two students were solving a quadratic equation in $x$. One student copied the constant term incorrectly and obtained the roots $3$ and $2$. The other student copied the constant term and the coefficient of $x^2$ correctly as $-6$ and $1$ respectively. What are the correct roots?

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