Product of real roots of the equation {t^2}{x | vedclass

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

Question

(C) Given the equation: ${t^2}{x^2} + |x| + 9 = 0$.For any real number $t$,the term ${t^2}{x^2} \ge 0$ and $|x| \ge 0$.Therefore,the expression ${t^2}

Vedclass · Accepted Answer

Does not exist

Vedclass · Answer

(C) Given the equation: ${t^2}{x^2} + |x| + 9 = 0$.For any real number $t$,the term ${t^2}{x^2} \ge 0$ and $|x| \ge 0$.Therefore,the expression ${t^2}{x^2} + |x| + 9$ must be at least $9$ for any real value of $x$.Since ${t^2}{x^2} + |x| + 9 \ge 9$,the expression can never equal $0$.Thus,the equation has no real roots.Consequently,the product of real roots does not exist.

Product of real roots of the equation ${t^2}{x^2} + |x| + 9 = 0$.

Similar Questions

If the roots of the equation $ax^2 + bx + c = 0$ are reciprocal to each other,then:

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$,then the equation whose roots are $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$ is:

The roots of the equation ${x^2} + 2\sqrt{3}x + 3 = 0$ are

The value of $k$ for which the equation $(K - 2)x^2 + 8x + K + 4 = 0$ has both roots real,distinct and negative is

$x=3$ is a solution of the equation $3x^{2} + (k-1)x + 9 = 0$ if $k$ has the value:

Vedclass Test Series

Exam Paper Generator

Online Exam Module