Question

If both roots of quadratic equation $(\alpha +1)x^2-2(1+3\alpha )x+1+8\alpha =0$ are real and distinict, the $\alpha$ be-

• A. -2
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is A -2

$(\alpha +1)x^2-2(1+3\alpha )x+1+8\alpha =0$

Here,

$a=\alpha +1$

$b=-2(1+3\alpha )$

$c=1+8\alpha$

If both roots are real and distinct,

$D>0$

$\Rightarrow b^2-4ac>0$

$\Rightarrow {(-2(1+3\alpha ))}^2-4(1+\alpha )(1+8\alpha )>0$

$\Rightarrow 4(1+9{\alpha }^2+6\alpha )-4(1+9\alpha +8{\alpha }^2)>0$

$\Rightarrow 4(1+9{\alpha }^2+6\alpha -1-9\alpha -8{\alpha }^2)>0$

$\Rightarrow {\alpha }^2-3\alpha >0$

$\Rightarrow \alpha (\alpha -3)>0$

$\Rightarrow \alpha \in (-\infty ,0)\cup (3,\infty )$

Hence among the given values, $\alpha$ will be $-2$.

Hence the correct answer is $\left(A\right)-2$.