The number of distinct real roots of (x - 2)^4 + (x + 2)^4 = 392 is

Steps to Solve

1. Rewrite the equation

Start with the equation:

$$(x - 2)^4 + (x + 2)^4 = 392$$

2. Introduce a substitution

Let:

$$u = (x - 2)^2 + (x + 2)^2$$

Then rewrite it as:

$$u = (x - 2)^2 + (x + 2)^2 = (x^2 - 4x + 4 + x^2 + 4x + 4) = 2x^2 + 8$$

3. Express the original equation in terms of $u$

Now, express the original equation using $u$:

$$(x - 2)^4 + (x + 2)^4 = u^2 - 16u + 64$$

Set it equal to 392:

$$u^2 - 16u + 64 = 392$$

4. Simplify the equation

Reorganize it:

$$u^2 - 16u - 328 = 0$$

5. Use the quadratic formula

Now apply the quadratic formula, $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

Here, $a = 1$, $b = -16$, and $c = -328$.

Calculate the discriminant:

$$D = b^2 - 4ac = (-16)^2 - 4(1)(-328) = 256 + 1312 = 1568$$

6. Solve for $u$

Now substitute the values into the quadratic formula:

$$u = \frac{16 \pm \sqrt{1568}}{2}$$

Calculating $\sqrt{1568}$ gives approximately $39.6$:

$$u = \frac{16 \pm 39.6}{2}$$

This yields two values:

1. $$u_1 = \frac{55.6}{2} \approx 27.8$$

2. $$u_2 = \frac{-23.6}{2} \approx -11.8$$ (not valid, since $u$ must be non-negative)

3. Back to $x$ and finding roots

Now, substitute back into $u = 2x^2 + 8$:

$$27.8 = 2x^2 + 8$$

Rearranging gives:

$$2x^2 = 19.8 \implies x^2 = 9.9 \implies x = \pm \sqrt{9.9}$$

Thus, we have two distinct real roots.