Q- If roots of quad. Equation $x^2-2ax+a^2+a-3=0$ are real and less than $3$ then,
a) $a<2$
b)$2<a<3$
c)$a>4$
In this ques., i used $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ and then if $a$ will be $1,2$ only then the root will be defined but if we use $3$ then there will be only one root but in ques. Roots are mentioned. Is the right.
Here roots are $a \pm\sqrt{3-a}$.
So roots to be real, $3-a>0 \implies a<3$.
And to satisfy second condition that roots be less than $3$, we see that
$a -\sqrt{3-a}<3$ iff $a<3$ and $a +\sqrt{3-a}<3$ iff $a<2$ ,
hence combining them all we get $a<2$.
The discriminant of the polynomial is $$ 4a^2-4(a^2+a-3)=12-4a $$ so you know that $12-4a\ge0$ and so $a\le 3$, which excludes (c).
For $a=0$, the equation is $$ x^2-3=0 $$ and the roots are less than $3$, which excludes (b).
Now, how can you completely verify the assert (a)? The largest root of the equation is $$ \frac{2a+\sqrt{12-4a}}{2}=a+\sqrt{3-a} $$ and the condition is thus $$ a+\sqrt{3-a}<3 $$ that's the same as $$ \sqrt{3-a}<(\sqrt{3-a})^2. $$ Since $a=3$ does not satisfy the inequality, we can reduce this to $$ 1<\sqrt{3-a} $$ or $$ 1<3-a $$ that's $a<2$.
