If, (a+1)x^2+2(a+1)x+a-2=0 then, for what parameter of ‘a’ the gi… - Vidyadip

MCQ

If, $(\text{a}+1)\text{x}^2+2(\text{a}+1)\text{x}+\text{a}-2=0$ then, for what parameter of ‘a’ the given equation have equal roots?

A
$\Big(-\infty,-1\Big)$
B
$\Big[-1,\infty\Big)$
C
$\Big(0,1\Big)$
✓
Not possible

✓

Answer

Correct option: D.

Not possible

For, equal roots, $\text{D}=0$
Where, $\text{D}=\text{b}^2-4\text{ac}$
In the equation, $(\text{a}+1)\text{x}^2+2(\text{a}+1)\text{x}+\text{a}-2=0$
$\text{D}=\Big[2(\text{a}+1)\Big]-4(\text{a}+1)(\text{a}-2)$
$=4\text{a}^2+4+8\text{a}-4(\text{a}^2-2\text{a}+\text{a}-2)$
$=4\text{a}^2+4+8\text{a}-4\text{a}^2+4\text{a}+8>0$
$\Rightarrow12\text{a}+12=0$
$\Rightarrow12\text{a}=-12$
$\Rightarrow\text{a}=-1$
So, from here it is clear that $\text{a}=-1$ is not possible because the equation is becoming linear.

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