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

If a root of the equation $x^2 + px + 12 = 0$ is $4$,while the roots of the equation $x^2 + px + q = 0$ are equal,then the value of $q$ will be:

If $x$ is real,then the value of $x^2 - 6x + 13$ will not be less than

Let $a \ne a_1 \ne 0,$ $f(x) = ax^2 + bx + c,$ $g(x) = a_1x^2 + b_1x + c_1,$ and $p(x) = f(x) - g(x).$ If $p(x) = 0$ only for $x = -1$ and $p(-2) = 2,$ then the value of $p(2)$ is

The sum of the non-real roots of $(p^2+p-3)(p^2+p-2)-12=0$ is

If $\alpha$ and $\beta$ are two complex roots of the equation $6x^6-25x^5+31x^4-31x^2+25x-6=0$,then $\alpha+\beta=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module