Roots(s) of the equation $9x^{2}-18|x|+5=0$ belonging to the domain of definition of the $f(x)=\log (x^{2}-x-2)$ is (are)
- $-\dfrac {5}{3}, -\dfrac {1}{3}$
- $\dfrac {5}{3}, \dfrac {1}{3}$
- $-\dfrac {5}{3}$
- $-\dfrac {1}{3}$
Let |x| = y. 9y^2 - 18y + 5 = 0. (3y-1)(3y-5) = 0. y = 1/3 or 5/3. |x| = 1/3 => x = +/- 1/3. |x| = 5/3 => x = +/- 5/3. Domain of log(x^2-x-2): x^2-x-2 > 0 => (x-2)(x+1) > 0. x > 2 or x < -1. Only -5/3 < -1.
Because of the absolute value, the equation becomes nine x squared minus eighteen x plus five equals zero for x greater than or equal to zero, yielding roots of one third and five thirds. For the domain of the logarithmic function, x squared minus x minus two must be greater than zero, meaning x is less than negative one or x is greater than two. The positive roots fail this domain condition, so we test negative x values. For x less than zero, the equation is nine x squared plus eighteen x plus five equals zero, giving roots of negative one third and negative five thirds; only negative five thirds is less than negative one and belongs to the domain.