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

We are asked to find the number of real roots for the equation:

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

This is a polynomial equation. To simplify it, we can use a substitution. Notice the terms \((1-x)\) and \((5-x)\). The average of 1 and 5 is \(\frac{1+5}{2} = 3\). Let's use the substitution \(y = x - 3\). This means \(x = y + 3\).

Substitute \(x = y + 3\) into the equation:

\((1 - (y + 3))^4 + (5 - (y + 3))^4 = 82\)

Simplify the terms inside the parentheses:

\((1 - y - 3)^4 + (5 - y - 3)^4 = 82\)

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

Note that \((-y - 2)^4 = (-(y + 2))^4 = (y + 2)^4\), because an even power makes the result positive. So the equation becomes:

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

Expanding the Equation using Binomial Theorem

We can expand \((y + 2)^4\) and \((2 - y)^4\) using the binomial theorem \((a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\).

For \((y + 2)^4\), \(a=y\) and \(b=2\):

\((y + 2)^4 = y^4 + 4y^3(2) + 6y^2(2^2) + 4y(2^3) + 2^4 = y^4 + 8y^3 + 24y^2 + 32y + 16\)

For \((2 - y)^4\), \(a=2\) and \(b=-y\):

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

Now, add the expansions together and set the sum equal to 82:

\((y^4 + 8y^3 + 24y^2 + 32y + 16) + (16 - 32y + 24y^2 - 8y^3 + y^4) = 82\)

Combine like terms:

• \(y^4\) terms: \(y^4 + y^4 = 2y^4\)
• \(y^3\) terms: \(8y^3 - 8y^3 = 0\)
• \(y^2\) terms: \(24y^2 + 24y^2 = 48y^2\)
• \(y\) terms: \(32y - 32y = 0\)
• Constant terms: \(16 + 16 = 32\)

The simplified equation in terms of \(y\) is:

\(2y^4 + 48y^2 + 32 = 82\)

Subtract 82 from both sides:

\(2y^4 + 48y^2 + 32 - 82 = 0\)

\(2y^4 + 48y^2 - 50 = 0\)

Divide the entire equation by 2:

\(y^4 + 24y^2 - 25 = 0\)

Solving for y

This is an equation that is quadratic in \(y^2\). Let \(z = y^2\). Substitute \(z\) into the equation:

\(z^2 + 24z - 25 = 0\)

Now we solve this quadratic equation for \(z\). We can factor it:

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

This gives two possible values for \(z\):

• \(z + 25 = 0 \implies z = -25\)
• \(z - 1 = 0 \implies z = 1\)

Now substitute back \(z = y^2\):

• Case 1: \(y^2 = -25\)
• Case 2: \(y^2 = 1\)

For Case 1, \(y^2 = -25\), the solutions for \(y\) are \(y = \pm \sqrt{-25} = \pm 5i\). These are complex (non-real) solutions for \(y\).

For Case 2, \(y^2 = 1\), the solutions for \(y\) are \(y = \pm \sqrt{1} = \pm 1\). These are real solutions for \(y\).

Finding the Real Roots for x

We used the substitution \(y = x - 3\), which means \(x = y + 3\). We need to find the real values of \(x\), so we only use the real values of \(y\) we found (\(y=1\) and \(y=-1\)).

• If \(y = 1\): \(x = 1 + 3 = 4\)
• If \(y = -1\): \(x = -1 + 3 = 2\)

So, the real roots of the original equation \((1 - x)^4 + (5 - x)^4 = 82\) are \(x = 4\) and \(x = 2\).

To verify, substitute these values back into the original equation:

• For \(x=4\): \((1 - 4)^4 + (5 - 4)^4 = (-3)^4 + (1)^4 = 81 + 1 = 82\). This is correct.
• For \(x=2\): \((1 - 2)^4 + (5 - 2)^4 = (-1)^4 + (3)^4 = 1 + 81 = 82\). This is correct.

Both values of \(x\) are real and satisfy the equation.

The equation has two real roots.

Revision Table: Key Steps in Finding Real Roots

Additional Information: Polynomial Equations and Roots

A polynomial equation of degree \(n\) can have at most \(n\) real roots. The given equation \((1 - x)^4 + (5 - x)^4 = 82\) expands to a polynomial of degree 4, so it can have at most 4 real roots.

When solving a polynomial equation, the roots can be real or complex. Real roots are values of the variable that are real numbers and satisfy the equation. Complex roots involve the imaginary unit \(i\) (where \(i^2 = -1\)).

For polynomial equations with real coefficients, complex roots always appear in conjugate pairs. In our case, the equation in \(y\) had complex roots \(\pm 5i\), which form a conjugate pair. These complex values of \(y\) lead to complex values of \(x\) through the substitution \(x = y+3\).

Using a substitution like \(y = x - \frac{a+b}{2}\) is a common technique to simplify equations of the form \((a-x)^n + (b-x)^n = C\). It works because the binomial expansions of \((y + k)^n\) and \((y - k)^n\) have cancelling terms (the odd powers of \(y\)) when added together if \(n\) is even.

For example, for \(n=4\):

\((y+k)^4 = y^4 + 4y^3k + 6y^2k^2 + 4yk^3 + k^4\)

\((y-k)^4 = y^4 - 4y^3k + 6y^2k^2 - 4yk^3 + k^4\)

Adding them: \((y+k)^4 + (y-k)^4 = 2y^4 + 12y^2k^2 + 2k^4\), which is an equation involving only even powers of \(y\), making it easier to solve by substituting \(z=y^2\).