If the roots of the equation x^2 - 2ax + a^2 | vedclass

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(A) Given equation is $x^2 - 2ax + a^2 + a - 3 = 0$.For real roots,the discriminant $D \ge 0$:$D = (-2a)^2 - 4(1)(a^2 + a - 3) \ge 0$$4a^2 - 4a^2 - 4a

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$a < 2$

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(A) Given equation is $x^2 - 2ax + a^2 + a - 3 = 0$.For real roots,the discriminant $D \ge 0$:$D = (-2a)^2 - 4(1)(a^2 + a - 3) \ge 0$$4a^2 - 4a^2 - 4a + 12 \ge 0$$-4a + 12 \ge 0 \Rightarrow a \le 3$.Let $f(x) = x^2 - 2ax + a^2 + a - 3$. Since the roots are less than $3$,the vertex of the parabola $x = -b/(2a) = a$ must be less than $3$,and $f(3) > 0$:$1$) $a $2$) $f(3) = 3^2 - 2a(3) + a^2 + a - 3 > 0$$9 - 6a + a^2 + a - 3 > 0$$a^2 - 5a + 6 > 0$$(a - 2)(a - 3) > 0$This implies $a  3$.Combining $a \le 3$,$a  3$),we get $a < 2$.

If the roots of the equation $x^2 - 2ax + a^2 + a - 3 = 0$ are real and less than $3$,then

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