Question

If $a\in (-1,1)$, then roots of the quadratic equation $(a-1)x^2+ax+\sqrt{1-a^2}=0$ are

• A. real
• B. imaginary
• C. both equal
• D. none of these

Answer 1

The correct option is A real

$(a-1)x^2+ax+\sqrt{1-a^2}=0$
$\Rightarrow -\sqrt{\left(\frac{1-a}{1+a}\right)}{x}^2+\frac{ax}{\sqrt{1-a^2}}+1=0$
Since, $a\in (-1,1)$, then put $a=\cos \theta$ to obtain
$-\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}{x}^2+x\cot \theta +1=0$
$\Rightarrow -\left|\tan \frac{\theta }{2}\right|x^2+x\cot \theta +1=0$
Now, $D={\cot }^2\theta +4\left|\tan \frac{\theta }{2}\right|>0$
Therefore, roots are real.
Ans: A