What is the domain of (sqrt(x))/(2x^2-5) ?

The domain of (sqrt(x))/(2x^2-5) is 0 sqrt(5/2)

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domain of (sqrt(x))/(2x^2-5)

domain

\frac{x}{2 x ^{2} - 5}

0 \leq x < \frac{5}{2} or x > \frac{5}{2}

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

x = \frac{5}{2}, x = - \frac{5}{2}

Combine real regions and undefined points for final function domain

0 \leq x < \frac{5}{2} or x > \frac{5}{2}

(\frac{x ^{2}}{4} - \frac{x}{2} - \frac{1}{4})^{2}

15

\frac{- 4 x + 20}{x ^{2} - 9 x + 20}

\frac{x ^{2} + 10 x + 21}{x ^{2} + 4 x - 21}

f (x) = 4 - \frac{x}{2}

What is the domain of (sqrt(x))/(2x^2-5) ?
The domain of (sqrt(x))/(2x^2-5) is 0<= x<sqrt(5/2)\lor x>sqrt(5/2)