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

Answer 1

The correct option is A $a<2$

Since the roots of the given equation are real
$\Rightarrow \therefore B^2-4AC\ge 0$
$\Rightarrow 4a^2-4(a^2+a-3)\ge 0$
$\Rightarrow -a^2-4(a^2+a-3)\ge 0$
$-a+3\ge 0ora\le 3$.........(1)
Since the roots are is less than 3,
so $f\left(3\right)>0$
$3^2-2a\left(3\right)+a^2+a-3>0$
$\Rightarrow a^2-5a+6>0or(a-2)(a-3)>0$
$\Rightarrow a<2ora>3$...........(2)
From (1) and (2), we have $a<2$