Question
If the roots of the equation $x^{2}-2 a x+a^{2}+a-3=0$ are real and less than 3 , then(A) $a<2$(B) $2 \leq a \leq 3$(B) $3 \leq a \leq 4$(D) $a>4$
Step 1
The discriminant of the equation $x^{2}-2 a x+a^{2}+a-3=0$ is $(-2a)^{2}-4(a^{2}+a-3)$, which simplifies to $4a^{2}-4a^{2}-4a+12= -4a+12$. So, we have $-4a+12 \geq 0$. Show more…
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