Find the real solutions, if any, of the following equation. Use

Ask by Whittaker Gonzalez. in the United States

Mar 31,2025

Question

Find the real solutions, if any, of the following equation. Use $$ 3-\frac{1}{x}-\frac{5}{x^{2}}=0 $$

UpStudy AI · Accepted Answer

1. We start with the equation  
   $$
   3 - \frac{1}{x} - \frac{5}{x^2} = 0.
   $$

2. Since the equation involves fractions with $$x$$ in the denominator, we multiply both sides by $$x^2$$ (noting that $$x \neq 0$$) to eliminate the fractions:
   $$
   x^2 \left(3 - \frac{1}{x} - \frac{5}{x^2}\right) = 0 \times x^2.
   $$
   This simplifies to:
   $$
   3x^2 - x - 5 = 0.
   $$

3. The resulting equation is a quadratic in $$x$$. We solve it using the quadratic formula:
   $$
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
   $$
   where for the equation $$3x^2 - x - 5 = 0$$, we have:
   $$
   a = 3, \quad b = -1, \quad c = -5.
   $$

4. Plug these values into the quadratic formula:
   $$
   x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-5)}}{2 \times 3} = \frac{1 \pm \sqrt{1 + 60}}{6}.
   $$

5. Simplify under the square root:
   $$
   x = \frac{1 \pm \sqrt{61}}{6}.
   $$

6. Thus, the two solutions are:
   $$
   x = \frac{1 + \sqrt{61}}{6} \quad \text{and} \quad x = \frac{1 - \sqrt{61}}{6}.
   $$

7. Both solutions are real since $$\sqrt{61}$$ is a real number and neither solution equals zero, which is consistent with our earlier domain restriction.

The real solutions to the equation are:  
$$
x = \frac{1 + \sqrt{61}}{6} \quad \text{and} \quad x = \frac{1 - \sqrt{61}}{6}.
$$
The real solutions are $$ x = \frac{1 + \sqrt{61}}{6} $$ and $$ x = \frac{1 - \sqrt{61}}{6} $$.

Find the real solutions, if any, of the following equation. Use

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