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If the roots of the equation $x^2-2 a x+a^2+a-3=0$ are real and less than 3 , then
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The correct answer is:
$a \lt 2$
Given equation is $x^2-2 a x+a^2+a-3=0$
If roots are real, then $D \geq 0$
$\Rightarrow 4 a^2-4\left(a^2+a-3\right) \geq 0 \Rightarrow-a+3 \geq 0$
$\Rightarrow a-3 \leq 0 \Rightarrow a \leq 3$
As roots are less than 3 , hence $f(3)\gt0$
$9-6 a+a^2+a-3\gt0 \Rightarrow a^2-5 a+6\gt0$
$\Rightarrow(a-2)(a-3)\gt0 \Rightarrow$ either $a \lt 2$ or $a\gt3$
Hence $a \lt 2$ satisfy all.
If roots are real, then $D \geq 0$
$\Rightarrow 4 a^2-4\left(a^2+a-3\right) \geq 0 \Rightarrow-a+3 \geq 0$
$\Rightarrow a-3 \leq 0 \Rightarrow a \leq 3$
As roots are less than 3 , hence $f(3)\gt0$
$9-6 a+a^2+a-3\gt0 \Rightarrow a^2-5 a+6\gt0$
$\Rightarrow(a-2)(a-3)\gt0 \Rightarrow$ either $a \lt 2$ or $a\gt3$
Hence $a \lt 2$ satisfy all.
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