Question

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

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

Answer 1

The correct option is A $a<2$

Given equation
$x^2-2ax+a^2+a-3=0\dots \left(1\right)$
$\Rightarrow (x-a{)}^2=3-a$
Since, $(x-a{)}^2\ge 0$
$\Rightarrow 3-a\ge 0$
$\Rightarrow a\le 3$

We have a formula for solving quadratic equation $a{x}^2+bx+c=0$ is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Now, the roots of equation $\left(1\right)$ are

$x=\frac{-(-2a)\pm \sqrt{(-2a{)}^2-\mathrm{4.1.}(a^2+a-3)}}{2.1}$

$x=a\pm \sqrt{3-a}$

Since, roots are less than 3.
$a+\sqrt{3-a}<3$

$\Rightarrow \sqrt{3-a}<3-a$

$\Rightarrow 3-a<9+a^2-6a$

$\Rightarrow a^2-5a+6>0$

$\Rightarrow (a-2)(a-3)>0$

$\Rightarrow a<2$ or $a>3$

Hence, $a<2$.