solve $$ 6 x^{4}+8 x^{2}=26 x^{3} $$

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To solve the equation:

$$ 6x^{4} + 8x^{2} = 26x^{3} $$

**Step 1: Rearrange the Equation**

First, bring all terms to one side to set the equation to zero:

$$ 6x^{4} - 26x^{3} + 8x^{2} = 0 $$

**Step 2: Factor Out Common Terms**

Notice that each term has an $$ x^{2} $$ factor:

$$ x^{2}(6x^{2} - 26x + 8) = 0 $$

This gives us two parts to solve:

1. $$ x^{2} = 0 $$
2. $$ 6x^{2} - 26x + 8 = 0 $$

**Step 3: Solve $$ x^{2} = 0 $$**

$$ x^{2} = 0 \implies x = 0 $$

This is a double root since the factor $$ x^{2} $$ appears.

**Step 4: Solve the Quadratic Equation**

Solve $$ 6x^{2} - 26x + 8 = 0 $$ using the quadratic formula:

$$ x = \frac{13 \pm \sqrt{(-13)^{2} - 4 \cdot 3 \cdot 4}}{2 \cdot 3} $$
$$ x = \frac{13 \pm \sqrt{169 - 48}}{6} $$
$$ x = \frac{13 \pm \sqrt{121}}{6} $$
$$ x = \frac{13 \pm 11}{6} $$

This yields two solutions:

1. $$ x = \frac{13 + 11}{6} = \frac{24}{6} = 4 $$
2. $$ x = \frac{13 - 11}{6} = \frac{2}{6} = \frac{1}{3} $$

**Final Solutions:**

- $$ x = 0 $$ (with multiplicity 2)
- $$ x = 4 $$
- $$ x = \frac{1}{3} $$

**Summary:**

The solutions to the equation $$ 6x^{4} + 8x^{2} = 26x^{3} $$ are:

$$ x = 0 \quad (\text{double root}), \quad x = 4, \quad \text{and} \quad x = \frac{1}{3} $$
The solutions are $$ x = 0 $$ (double root), $$ x = 4 $$, and $$ x = \frac{1}{3} $$.

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