Question

One of the roots of the equation \((x+1)^4 + 81 = 0\) is

• A. \(3(\frac{1+i}{\sqrt{2}})\)
• B. \(\frac{-3+\sqrt{2}+3i}{\sqrt{2}}\)
• C. \(-(\frac{3+\sqrt{2}+i}{\sqrt{2}})\)
• D. \(-(\frac{3+3i}{\sqrt{2}})\)

Hint:

When the roots you calculate for a polynomial equation don't match any of the given options, double-check the problem statement for simple sign errors. A common typo is having `(x+a)` when `(x-a)` was intended, or vice-versa. Testing the modified equation can quickly lead to the correct option.

Answer 1

The Correct Option is B

Let \(y = x+1\). The equation transforms to \(y^4 + 81 = 0\), or \(y^4 = -81\).
We need to find the fourth roots of -81. First, write -81 in polar form.
\(-81 = 81(\cos(\pi) + i\sin(\pi))\).
The fourth roots are given by \(y_k = \sqrt[4]{81} \left[ \cos\left(\frac{\pi + 2k\pi}{4}\right) + i\sin\left(\frac{\pi + 2k\pi}{4}\right) \right]\) for \(k = 0, 1, 2, 3\).
The principal root is \( \sqrt[4]{81} = 3 \).
For \(k=0: y_0 = 3(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4})) = 3(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) = \frac{3+3i}{\sqrt{2}}\).
For \(k=1: y_1 = 3(\cos(\frac{3\pi}{4}) + i\sin(\frac{3\pi}{4})) = 3(-\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) = \frac{-3+3i}{\sqrt{2}}\).
For \(k=2: y_2 = 3(\cos(\frac{5\pi}{4}) + i\sin(\frac{5\pi}{4})) = 3(-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) = \frac{-3-3i}{\sqrt{2}}\).
For \(k=3: y_3 = 3(\cos(\frac{7\pi}{4}) + i\sin(\frac{7\pi}{4})) = 3(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) = \frac{3-3i}{\sqrt{2}}\).
The roots of the original equation are \(x = y-1\). However, none of the \(y_k-1\) match the options.
This suggests a likely typo in the question, and it was intended to be \((x-1)^4 + 81 = 0\). Let's solve for this case.
If the equation is \((x-1)^4 = -81\), then \(x-1 = y\), so \(x = y+1\).
Let's find the corresponding values of \(x\).
\(x_1 = y_1+1 = \frac{-3+3i}{\sqrt{2}} + 1 = \frac{-3+3i+\sqrt{2}}{\sqrt{2}} = \frac{-3+\sqrt{2}+3i}{\sqrt{2}}\).
This value exactly matches option (B).