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

What is the minimum value of the expression $x^2 + 4y^2 + 3z^2 - 2x - 12y - 6z + 14$?

Let $b$ be a non-zero real number. Suppose the quadratic equation $2x^2 + bx + \frac{1}{b} = 0$ has two distinct real roots. Then:

Which of the following is not a root of $f(x) = x^3 - 11x^2 + 36x - 36$?

The exact set of values of $a$ for which the equation ${x^3}(x + 1) = 2(x + a)(x + 2a)$ has four real solutions is:

The values of $a$ for which the expression $(a^2 - 1)x^2 + 2(a - 1)x + 2$ is positive for all $x \in \mathbb{R}$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module