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

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(B) Given the equation: $9x^{2}-18|x|+5=0$.Since $x^{2} = |x|^{2}$,we can rewrite the equation as $9|x|^{2}-18|x|+5=0$.Let $|x| = t$. Then the equatio

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$\frac{25}{81}$

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(B) Given the equation: $9x^{2}-18|x|+5=0$.Since $x^{2} = |x|^{2}$,we can rewrite the equation as $9|x|^{2}-18|x|+5=0$.Let $|x| = t$. Then the equation becomes $9t^{2}-18t+5=0$.Factoring the quadratic equation: $9t^{2}-15t-3t+5=0$.$3t(3t-5)-1(3t-5)=0$.$(3t-1)(3t-5)=0$.So,$t = \frac{1}{3}$ or $t = \frac{5}{3}$.Since $|x| = t$,we have $|x| = \frac{1}{3}$ and $|x| = \frac{5}{3}$.This gives the roots: $x = \pm \frac{1}{3}$ and $x = \pm \frac{5}{3}$.The roots are $\frac{1}{3}, -\frac{1}{3}, \frac{5}{3}, -\frac{5}{3}$.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

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