Question

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

• A. $a<2$
• B. $a\le a\le 3$
• C. $3<a\le 4$
• D. $a>2$

Answer 1

The correct option is A $a<2$

Solution : $x^2-2ax+a^2+a-3=0$
$D=4a^2-4(a^2+a-3)\ge 0$
$a-3\le 0$
$a\le 3$
$f\left(3\right)=9-6a+a^2+a-3>0$
$a^2-5a+6>0$
$(a-2)(a-3)>0$
$a<2ora>3$
$\frac{2a}{a}<3$
$a<3$
$(a<2)$