Let S be the set of all non-zero real numbers | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For the quadratic equation $\alpha x^2 - x + \alpha = 0$ to have two distinct real roots,the discriminant $D > 0$.$D = (-1)^2 - 4(\alpha)(\alpha)

Vedclass · Accepted Answer

$(A, D)$

Vedclass · Answer

(D) For the quadratic equation $\alpha x^2 - x + \alpha = 0$ to have two distinct real roots,the discriminant $D > 0$.$D = (-1)^2 - 4(\alpha)(\alpha) = 1 - 4\alpha^2 > 0$ $\Rightarrow \alpha^2 Given $|x_1 - x_2| Using $(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2$,we get $\left(\frac{1}{\alpha}\right)^2 - 4(1) $\frac{1}{\alpha^2} - 4  \frac{1}{5}$.So,$|\alpha| > \frac{1}{\sqrt{5}}$,which means $\alpha \in \left(-\infty, -\frac{1}{\sqrt{5}}\right) \cup \left(\frac{1}{\sqrt{5}}, \infty\right)$.Combining the conditions,$\alpha \in \left(-\frac{1}{2}, -\frac{1}{\sqrt{5}}\right) \cup \left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)$.Thus,the intervals $(A)$ and $(D)$ are subsets of $S$.

Similar Questions

The number of distinct real roots of the equation $x^{5}(x^{3}-x^{2}-x+1)+x(3x^{3}-4x^{2}-2x+4)-1=0$ is

For the equation $|x^2| + |x| - 6 = 0$,the roots are

If the roots of $x^4+x^3-4x^2+x+1=0$ are diminished by $\alpha$ or $\beta$,then the equation with the diminished roots does not contain the $x^2$ term. Then $12(\alpha-\beta)^2=$

The polynomial $x^2-6x+12 \in \mathbb{Q}[x]$ is

If $A$ and $G$ are arithmetic and geometric means and ${x^2} - 2Ax + {G^2} = 0$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module