The product of the roots of the equation 9x^{ | vedclass

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

Question

(B) Given equation: $9x^{2}-18|x|+5=0$Since $x^{2} = |x|^{2}$,the equation becomes $9|x|^{2}-18|x|+5=0$.Let $t = |x|$,then $9t^{2}-18t+5=0$.Factoring

Vedclass · Accepted Answer

$\frac{25}{81}$

Vedclass · Answer

(B) Given equation: $9x^{2}-18|x|+5=0$Since $x^{2} = |x|^{2}$,the equation becomes $9|x|^{2}-18|x|+5=0$.Let $t = |x|$,then $9t^{2}-18t+5=0$.Factoring the quadratic: $9t^{2}-15t-3t+5=0$.$3t(3t-5)-1(3t-5)=0$.$(3t-1)(3t-5)=0$.So,$|x| = \frac{1}{3}$ or $|x| = \frac{5}{3}$.The roots are $x = \frac{1}{3}, -\frac{1}{3}, \frac{5}{3}, -\frac{5}{3}$.The product of the roots is $(\frac{1}{3}) 	imes (-\frac{1}{3}) 	imes (\frac{5}{3}) 	imes (-\frac{5}{3}) = \frac{25}{81}$.

The product of the roots of the equation $9x^{2}-18|x|+5=0$ is

Similar Questions

The roots of the quadratic equation $(a + b - 2c)x^2 - (2a - b - c)x + (a - 2b + c) = 0$ are

If $(x^2+5x+5)^{x+5}=1$,then the number of integers satisfying this equation is

If the product of roots of the equation ${x^2} - 3kx + 2{e^{2\log k}} - 1 = 0$ is $7$,then its roots will be real when:

The equation $\frac{3}{x - a^3} + \frac{5}{x - a^5} + \frac{7}{x - a^7} = 0$,where $a > 1$,has:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module