If 8 and 2 are the roots of {x^2} + ax + beta | vedclass

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(D) Given that $8$ and $2$ are the roots of ${x^2} + ax + \beta = 0$. Using the relation between roots and coefficients,the sum of roots is $8 + 2 = 1

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$9, 1$

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(D) Given that $8$ and $2$ are the roots of ${x^2} + ax + \beta = 0$. Using the relation between roots and coefficients,the sum of roots is $8 + 2 = 10 = -a$,so $a = -10$. The product of roots is $8 	imes 2 = 16 = \beta$. Given that $3$ and $3$ are the roots of ${x^2} + \alpha x + b = 0$. The sum of roots is $3 + 3 = 6 = -\alpha$,so $\alpha = -6$. The product of roots is $3 	imes 3 = 9 = b$. Now,substituting $a = -10$ and $b = 9$ into the equation ${x^2} + ax + b = 0$,we get ${x^2} - 10x + 9 = 0$. Factoring the quadratic equation: $(x - 1)(x - 9) = 0$. Thus,the roots are $x = 1$ and $x = 9$.

If $8$ and $2$ are the roots of ${x^2} + ax + \beta = 0$ and $3$ and $3$ are the roots of ${x^2} + \alpha x + b = 0$,then the roots of ${x^2} + ax + b = 0$ are

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