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 < a $\le$ 4
• D. a > 4

Answer 1

Answer: (A)

Given equation
$x^2-2ax+a^2+a-3=\mathrm{0....}$ (1)
$\Rightarrow (x-a{)}^2=3-a$
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 (1) are
$x=\frac{-(-2a)\pm \sqrt{(-2a{)}^2-4\cdot 1\cdot \left(a^2+a-3\right)}}{2\cdot 1}$
$x=a\pm \sqrt{3-a}$
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, $a<2$.