Solving the Equation (1 - x)$^4$ + (5 - x)$^4$ = 82 and Finding the Sum of Roots

The problem asks us to find the sum of all the roots of the given equation:

$$ (1 - x)^4 + (5 - x)^4 = 82 $$

This equation is symmetric around a certain point. To simplify it, we can make a substitution that exploits this symmetry. The midpoint of 1 and 5 is $$(1+5)/2 = 3$$. Let's substitute $$ y = x - 3$$. Then $$ x = y + 3$$.

Substituting $$ x = y + 3 $$ into the equation:

• $$ 1 - x = 1 - (y + 3) = 1 - y - 3 = -y - 2 = -(y + 2) $$
• $$ 5 - x = 5 - (y + 3) = 5 - y - 3 = -y + 2 = 2 - y $$

The equation becomes:

$$ (-(y + 2))^4 + (2 - y)^4 = 82 $$

Since $$ (-a)^4 = a^4 $$, this simplifies to:

$$ (y + 2)^4 + (2 - y)^4 = 82 $$

Now, let's expand the terms using the binomial theorem or Pascal's triangle (coefficients for power 4 are 1, 4, 6, 4, 1):

• $$ (y + 2)^4 = y^4 + 4(y^3)(2) + 6(y^2)(2^2) + 4(y)(2^3) + 2^4 = y^4 + 8y^3 + 24y^2 + 32y + 16 $$
• $$ (2 - y)^4 = (-(y - 2))^4 = (y - 2)^4 = y^4 + 4(y^3)(-2) + 6(y^2)(-2)^2 + 4(y)(-2)^3 + (-2)^4 $$
• $$ (y - 2)^4 = y^4 - 8y^3 + 24y^2 - 32y + 16 $$

Adding these two expansions:

$$ (y^4 + 8y^3 + 24y^2 + 32y + 16) + (y^4 - 8y^3 + 24y^2 - 32y + 16) = 2y^4 + 48y^2 + 32 $$

So the equation in terms of $$ y $$ is:

$$ 2y^4 + 48y^2 + 32 = 82 $$

Subtract 82 from both sides:

$$ 2y^4 + 48y^2 - 50 = 0 $$

Divide by 2:

$$ y^4 + 24y^2 - 25 = 0 $$

This is a biquadratic equation. We can solve it by treating it as a quadratic equation in $$ y^2 $$. Let $$ z = y^2 $$.

$$ z^2 + 24z - 25 = 0 $$

We can factor this quadratic equation:

$$ (z + 25)(z - 1) = 0 $$

This gives two possible values for $$ z $$:

• $$ z + 25 = 0 \implies z = -25 $$
• $$ z - 1 = 0 \implies z = 1 $$

Now, substitute back $$ y^2 = z $$.

• Case 1: $$ y^2 = -25 $$
  Taking the square root, $$ y = \pm \sqrt{-25} = \pm 5i $$.
• Case 2: $$ y^2 = 1 $$
  Taking the square root, $$ y = \pm \sqrt{1} = \pm 1 $$.

So, the possible values for $$ y $$ are $$ 1, -1, 5i, -5i $$.

Finally, substitute back $$ x = y + 3 $$ to find the roots of the original equation:

• If $$ y = 1$$, $$ x = 1 + 3 = 4 $$
• If $$ y = -1$$, $$ x = -1 + 3 = 2 $$
• If $$ y = 5i$$, $$ x = 3 + 5i $$
• If $$ y = -5i$$, $$ x = 3 - 5i $$

The roots of the equation are $$ 4, 2, 3 + 5i, 3 - 5i $$.

Calculating the Sum of the Roots

To find the sum of all the roots, we add them together:

Sum $$ = 4 + 2 + (3 + 5i) + (3 - 5i) $$

Sum $$ = 4 + 2 + 3 + 5i + 3 - 5i $$

The imaginary parts $$ +5i $$ and $$ -5i $$ cancel each other out.

Sum $$ = 4 + 2 + 3 + 3 $$

Sum $$ = 12 $$

The sum of all the roots of the equation is 12.

Revision Table: Key Steps in Solving the Equation

Additional Information on Sum of Roots and Complex Conjugates

For a general polynomial equation of degree $$ n $$, say $$ a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 = 0 $$, the sum of the roots is given by $$ - \frac{a_{n-1}}{a_n} $$.

In our case, the original equation was $$(1-x)^4 + (5-x)^4 = 82$$. Expanding this fully would give a polynomial in terms of $$ x $$. Let's look at the highest degree terms:

• $$(1-x)^4 = (-(x-1))^4 = (x-1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1$$
• $$(5-x)^4 = (-(x-5))^4 = (x-5)^4 = x^4 - 4x^3(5) + 6x^2(5^2) - 4x(5^3) + 5^4 $$
• $$ = x^4 - 20x^3 + 150x^2 - 500x + 625 $$

Adding these two expansions:

$$(x^4 - 4x^3 + 6x^2 - 4x + 1) + (x^4 - 20x^3 + 150x^2 - 500x + 625) = 2x^4 - 24x^3 + 156x^2 - 504x + 626$$

So the equation is $$ 2x^4 - 24x^3 + 156x^2 - 504x + 626 = 82 $$.

Subtracting 82 from both sides:

$$ 2x^4 - 24x^3 + 156x^2 - 504x + 544 = 0 $$

Comparing this to the general form $$ a_n x^n + a_{n-1} x^{n-1} + \dots = 0 $$, we have $$ n=4 $$, $$ a_4 = 2 $$, and $$ a_3 = -24 $$.

The sum of the roots is $$ - \frac{a_3}{a_4} = - \frac{-24}{2} = \frac{24}{2} = 12 $$.

This confirms the result obtained by finding the individual roots. This method using Vieta's formulas is generally faster for finding the sum or product of roots without finding the roots themselves, provided the polynomial is in standard form.

Also notice the roots $$ 3+5i $$ and $$ 3-5i $$ are complex conjugates. For polynomial equations with real coefficients, complex roots always appear in conjugate pairs. The sum of a complex conjugate pair $$(a + bi) + (a - bi) = 2a$$, which is a real number. The presence of complex conjugate roots does not affect the sum of roots calculation using Vieta's formulas.