3x^4 - 14x^2 = 5

Steps to Solve

1. Rearrange the equation to standard form

Start with the original equation:

$$ 3x^4 - 14x^2 = 5 $$

Subtract 5 from both sides to set the equation to 0:

$$ 3x^4 - 14x^2 - 5 = 0 $$

2. Substitute variables

Let ( y = x^2 ) to simplify the polynomial. Then the equation becomes:

$$ 3y^2 - 14y - 5 = 0 $$

3. Use the quadratic formula

We will apply the quadratic formula, given by:

$$ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

In this case, ( a = 3 ), ( b = -14 ), and ( c = -5 ).

4. Calculate the discriminant

Find the discriminant:

$$ b^2 - 4ac = (-14)^2 - 4 \cdot 3 \cdot (-5) $$

Calculate:

$$ 196 + 60 = 256 $$

5. Solve for y

Now plug in the values into the quadratic formula:

$$ y = \frac{-(-14) \pm \sqrt{256}}{2 \cdot 3} $$

Calculate:

$$ y = \frac{14 \pm 16}{6} $$

Which gives:

$$ y_1 = \frac{30}{6} = 5 \quad \text{and} \quad y_2 = \frac{-2}{6} = -\frac{1}{3} $$

6. Back substitute for x

Recall ( y = x^2 ). Therefore, we have:

For ( y_1 = 5 ):

$$ x^2 = 5 \implies x = \sqrt{5} \text{ or } x = -\sqrt{5} $$

For ( y_2 = -\frac{1}{3} ):

Since ( x^2 ) cannot be negative, this case does not provide any real solutions.

7. Final solutions

Thus, the real solutions for the original polynomial equation are:

$$ x = \sqrt{5} \quad \text{and} \quad x = -\sqrt{5} $$