Find all real solutions to the equation:

Ask by Dawson Evans. in the United States

Jan 25,2025

Question

Find all real solutions to the equation: $$ 2 x^{3}-25=-x^{2}+50 x $$ $$ x=\square $$

UpStudy AI · Accepted Answer

To solve the equation:

$$
2x^{3} - 25 = -x^{2} + 50x
$$

First, move all terms to one side of the equation:

$$
2x^{3} + x^{2} - 50x - 25 = 0
$$

We now have a cubic equation:

$$
2x^3 + x^2 - 50x - 25 = 0
$$

**Step 1: Find Rational Roots**

Using the Rational Root Theorem, possible rational roots are factors of the constant term divided by factors of the leading coefficient. The possible rational roots are:

$$
\pm 1, \pm \frac{1}{2}, \pm 5, \pm \frac{5}{2}, \pm 25, \pm \frac{25}{2}
$$

Testing $$ x = 5 $$:

$$
2(5)^3 + (5)^2 - 50(5) - 25 = 250 + 25 - 250 - 25 = 0
$$

So, $$ x = 5 $$ is a root.

**Step 2: Factor the Polynomial**

Factor out $$ (x - 5) $$ from the cubic equation:

$$
2x^3 + x^2 - 50x - 25 = (x - 5)(2x^2 + 11x + 5)
$$

**Step 3: Solve the Quadratic Equation**

Now, solve the quadratic equation:

$$
2x^2 + 11x + 5 = 0
$$

Using the quadratic formula:

$$
x = \frac{-11 \pm \sqrt{121 - 40}}{4} = \frac{-11 \pm \sqrt{81}}{4} = \frac{-11 \pm 9}{4}
$$

This gives two solutions:

$$
x = \frac{-11 + 9}{4} = -\frac{1}{2} \quad \text{and} \quad x = \frac{-11 - 9}{4} = -5
$$

**Final Solutions**

All real solutions to the equation are:

$$
x = 5, \quad x = -5, \quad \text{and} \quad x = -\frac{1}{2}
$$

**Answer:**

$$x = 5,\; x = -5,\; \text{and}\; x = -\dfrac{1}{2}$$
The real solutions are $$ x = 5 $$, $$ x = -5 $$, and $$ x = -\frac{1}{2} $$.

Find all real solutions to the equation:

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