If S = {m in mathbb{R} : x^2 - 2(1 + 3m)x + 7 | vedclass

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

Question

(D) The given quadratic equation is $x^2 - 2(1 + 3m)x + 7(3 + 2m) = 0$.
For the roots to be distinct,the discriminant $D$ must be greater than $0$.

Vedclass · Accepted Answer

infinite

Vedclass · Answer

(D) The given quadratic equation is $x^2 - 2(1 + 3m)x + 7(3 + 2m) = 0$.
For the roots to be distinct,the discriminant $D$ must be greater than $0$.
$D = b^2 - 4ac > 0$
$[-2(1 + 3m)]^2 - 4(1)(7(3 + 2m)) > 0$
$4(1 + 9m^2 + 6m) - 28(3 + 2m) > 0$
$4 + 36m^2 + 24m - 84 - 56m > 0$
$36m^2 - 32m - 80 > 0$
Dividing by $4$:
$9m^2 - 8m - 20 > 0$
Factoring the quadratic expression:
$(9m + 10)(m - 2) > 0$
The roots of the equation $9m^2 - 8m - 20 = 0$ are $m = 2$ and $m = -\frac{10}{9}$.
Thus,the inequality holds for $m \in (-\infty, -\frac{10}{9}) \cup (2, \infty)$.
Since $S$ is a subset of the set of real numbers $\mathbb{R}$ and the interval $(-\infty, -\frac{10}{9}) \cup (2, \infty)$ contains infinitely many real numbers,the number of elements in $S$ is infinite.

If $S = {m \in \mathbb{R} : x^2 - 2(1 + 3m)x + 7(3 + 2m) = 0}$ has distinct roots $\}$,then the number of elements in $S$ is

Similar Questions

The roots of $|x - 2|^2 + |x - 2| - 6 = 0$ are

If the quadratic equation $(\lambda + 2)x^2 - 3\lambda x + 4\lambda = 0, \lambda \neq -2$, has two positive roots, then the number of possible integral values of $\lambda$ is:

What is the nature of the roots of the equation $x^2 + x = 2(x - 1)$?

The smallest positive integral value of $a$, for which all the roots of $x^4 - ax^2 + 9 = 0$ are real and distinct, is equal to

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+3x^2-10x-24=0$,and $\alpha(\beta+\gamma), \beta(\gamma+\alpha), \gamma(\alpha+\beta)$ are the roots of the equation $x^3+px^2+qx+r=0$,then $q=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module