# If the two roots of the equation, are real and distinct, then the set of all values of ‘a’ is : Option 1) Option 2) Option 3) Option 4)

As we learnt in

Condition for Real and distinct roots of Quadratic Equation -

$D= b^{2}-4ac> 0$

- wherein

$ax^{2}+bx+c= 0$

$(a-1)(x^{4}+x^{2}+1)+(a+1)(x^{2}+x+1)^{2}=0$

$\Rightarrow\ \; (a-1)(x^{2}+2x^{2}+1-x^{2})+(a+1)(x^{2}+x+1)^{2}=0$

$\Rightarrow\ \; (a-1)((x^{2}+1)^{2}-x^{2})+(a+1)(x^{2}+x+1)^{2}=0$

$\Rightarrow\ \; (a-1)(x^{2}+1+x)(x^{2}+1-x)+(a+1)(x^{2}+x+1)^{2}=0$

$\Rightarrow\ \; (x^{2}+x+1)((a-1)(x^{2}-x+1))+(a+1)(x^{2}+x+1)=0$

$\Rightarrow\ \; x^{2}+x+1 \neq0$

So $(a-1)(x^{2}-x+1)+(a+1)(x^{2}+x+1)=0$

$\Rightarrow\ \; 2ax^{2}+2x+2a=0$

$\Rightarrow\ \; ax^{2}+x+a=0$

So for real value of x  $D\geq0$

$\therefore\ \;1-4.a.a\geq0$

$1\geq4a^{2}\ \; \Rightarrow\ \;4a^{2}\leq1$

$\therefore\ \; \frac{-1}{2}\leq a \leq \frac{1}{2}$

But $a\neq0$

So $a \epsilon \left(\frac{-1}{2},0 \right )\bigcup \left(0,\frac{1}{\sqrt{2}} \right )$

Correct option is 3.

Option 1)

This is an incorrect option.

Option 2)

This is an incorrect option.

Option 3)

This is the correct option.

Option 4)

This is an incorrect option.

N

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