If the roots of the equation x^2+2ax+b=0 are | vedclass

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(A) Let the roots of the equation $x^2+2ax+b=0$ be $\alpha$ and $\beta$. Since the roots are real and distinct,the discriminant $D > 0$. $D = (2a)^2 -

Vedclass · Accepted Answer

$(a^2-m^2, a^2)$

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(A) Let the roots of the equation $x^2+2ax+b=0$ be $\alpha$ and $\beta$. Since the roots are real and distinct,the discriminant $D > 0$. $D = (2a)^2 - 4(1)(b) = 4a^2 - 4b > 0 \implies a^2 > b$ or $b The roots are given by $\alpha, \beta = \frac{-2a \pm \sqrt{4a^2-4b}}{2} = -a \pm \sqrt{a^2-b}$. The difference between the roots is $|\alpha - \beta| = |2\sqrt{a^2-b}|$. Given that the roots differ at most by $2m$,we have $2\sqrt{a^2-b} \le 2m$. $\sqrt{a^2-b} \le m \implies a^2-b \le m^2 \implies b \ge a^2-m^2$. Combining the conditions $b Thus,the correct option is $A$.

If the roots of the equation $x^2+2ax+b=0$ are real,distinct and differ at most by $2m$,then $b$ lies in the interval

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