Question

Which of the following is/are true regarding the quadratic equation $(a+1)x^2+2(a+1)x+a-2=0$?

• A. If $a>(-1)$ then this equation has two distinct roots.
• B. If $a=-1$, then this equation has two equal roots
• C. If $a<(-1)$, then this equation has no real roots.
• D. All of the above

Answer 1

Answer: (A), (C)

If $a<(-1)$, then this equation has no real roots.

For quadratic equation $a{x}^2+bx+c=0,a\ne 0$ Nature of roots is given by discriminant $D=b^2-4ac$ as $\left(i\right)D>0\Rightarrow \text{Real and distinct roots}\left(ii\right)D=0\Rightarrow \text{Real and equal roots}\left(iii\right)D<0\Rightarrow \text{Non real or imaginary roots}$

Given equation is $(a+1)x^2+2(a+1)x+a-2=0$
Since this is a quadratic equation,
$(a+1)\ne 0\Rightarrow a\ne -1$.
Now, $D=4(a+1{)}^2-4(a+1\left)\right(a-2)$
$\Rightarrow D=4(a+1)(a+1-a+2)$
$\Rightarrow D=12(a+1)$

If $a>(-1)$, then $D>0$, this equation has two distinct roots.
If $a<(-1)$, then $D<0$, this equation has no real roots.
If $D=0$, $a$ has to be $-1$. But since $a\ne -1$, this equation cannot have two equal roots