Thank you for reporting, we will resolve it shortly
Q.
If the roots of the equation $ x^2 - 2ax + a^2 + a - 3 = 0 $ are real and less than 3, then
IIT JEEIIT JEE 1999Complex Numbers and Quadratic Equations
Solution:
Given equation
$x^{2}-2 a x+a^{2}+a-3=0 ....$ (1)
$\Rightarrow(x-a)^{2}=3-a $
Since, $(x-a)^{2} \geq 0 $
$\Rightarrow 3-a \geq 0 $
$\Rightarrow a \leq 3 $
We have a formula for solving quadratic equation $a x^{2}+b x+c=0$ is
$ x =\frac{- b \pm \sqrt{ b ^{2}-4 ac }}{2 a } $
Now, the roots of equation (1) are
$ x =\frac{-(-2 a ) \pm \sqrt{(-2 a )^{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 $3$ .
$a+\sqrt{3-a} < 3$
$\Rightarrow \sqrt{3- a } < 3- a$
$\Rightarrow 3- a < 9+ a ^{2}-6 a$
$\Rightarrow a ^{2}-5 a +6>0$
$\Rightarrow(a-2)(a-3)>0$
$\Rightarrow a < 2 $ or $ a >3$
Hence, $a < 2$.