Understanding the Problem

We are given an expression for x as \(x = 97 + 56\sqrt{3}\). We need to find the value of the expression \(\sqrt[4]{x}+\frac{1}{\sqrt[4]{x}}\). This involves finding the fourth root of x and its reciprocal, and then summing them up.

To find the fourth root, we can first find the square root of x, and then find the square root of that result. This means we need to simplify the radical expression \(\sqrt{97 + 56\sqrt{3}}\) and then \(\sqrt{\sqrt{97 + 56\sqrt{3}}}\).

Step-by-Step Calculation of the Fourth Root

Let's start by simplifying \(\sqrt{97 + 56\sqrt{3}}\). We look for two numbers whose sum of squares is 97 and whose product, when multiplied by 2 and the square root of 3, gives \(56\sqrt{3}\). This form suggests we are looking for a perfect square of the form \((a + b\sqrt{3})^2 = a^2 + (b\sqrt{3})^2 + 2 \cdot a \cdot b\sqrt{3} = a^2 + 3b^2 + 2ab\sqrt{3}\).

Comparing \(97 + 56\sqrt{3}\) with \(a^2 + 3b^2 + 2ab\sqrt{3}\):

• We have \(2ab = 56\), which means \(ab = 28\).
• We also have \(a^2 + 3b^2 = 97\).

Let's consider pairs of factors of 28 for \(a\) and \(b\): (1, 28), (2, 14), (4, 7), (7, 4), (14, 2), (28, 1).

Trying these pairs in \(a^2 + 3b^2 = 97\):

• If \(a=1, b=28\): \(1^2 + 3(28^2) = 1 + 3(784) = 1 + 2352 \neq 97\).
• If \(a=2, b=14\): \(2^2 + 3(14^2) = 4 + 3(196) = 4 + 588 \neq 97\).
• If \(a=4, b=7\): \(4^2 + 3(7^2) = 16 + 3(49) = 16 + 147 \neq 97\).
• If \(a=7, b=4\): \(7^2 + 3(4^2) = 49 + 3(16) = 49 + 48 = 97\). This pair works!

So, \(97 + 56\sqrt{3} = (7 + 4\sqrt{3})^2\).

Therefore, \(\sqrt{x} = \sqrt{(7 + 4\sqrt{3})^2} = 7 + 4\sqrt{3}\) (since \(7 + 4\sqrt{3}\) is positive).

Finding the Fourth Root of x

Now we need to find the square root of \(7 + 4\sqrt{3}\). Again, we look for a perfect square of the form \((c + d\sqrt{3})^2 = c^2 + 3d^2 + 2cd\sqrt{3}\).

Comparing \(7 + 4\sqrt{3}\) with \(c^2 + 3d^2 + 2cd\sqrt{3}\):

• We have \(2cd = 4\), which means \(cd = 2\).
• We also have \(c^2 + 3d^2 = 7\).

Let's consider pairs of factors of 2 for \(c\) and \(d\): (1, 2), (2, 1).

Trying these pairs in \(c^2 + 3d^2 = 7\):

• If \(c=1, d=2\): \(1^2 + 3(2^2) = 1 + 3(4) = 1 + 12 = 13 \neq 7\).
• If \(c=2, d=1\): \(2^2 + 3(1^2) = 4 + 3(1) = 4 + 3 = 7\). This pair works!

So, \(7 + 4\sqrt{3} = (2 + 1\sqrt{3})^2 = (2 + \sqrt{3})^2\).

Therefore, \(\sqrt[4]{x} = \sqrt{\sqrt{x}} = \sqrt{7 + 4\sqrt{3}} = \sqrt{(2 + \sqrt{3})^2} = 2 + \sqrt{3}\) (since \(2 + \sqrt{3}\) is positive).

Calculating the Reciprocal

Next, we need to find the reciprocal of \(\sqrt[4]{x}\), which is \(\frac{1}{\sqrt[4]{x}} = \frac{1}{2 + \sqrt{3}}\). To simplify this, we can rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is \(2 - \sqrt{3}\).

\(\frac{1}{2 + \sqrt{3}} = \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{(2)^2 - (\sqrt{3})^2} = \frac{2 - \sqrt{3}}{4 - 3} = \frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}\).

Finding the Final Value

Now we can find the value of the expression \(\sqrt[4]{x}+\frac{1}{\sqrt[4]{x}}\).

\(\sqrt[4]{x}+\frac{1}{\sqrt[4]{x}} = (2 + \sqrt{3}) + (2 - \sqrt{3})\)

\(\sqrt[4]{x}+\frac{1}{\sqrt[4]{x}} = 2 + \sqrt{3} + 2 - \sqrt{3}\)

\(\sqrt[4]{x}+\frac{1}{\sqrt[4]{x}} = 2 + 2 + (\sqrt{3} - \sqrt{3})\)

\(\sqrt[4]{x}+\frac{1}{\sqrt[4]{x}} = 4 + 0\)

\(\sqrt[4]{x}+\frac{1}{\sqrt[4]{x}} = 4\)

The value of the expression is 4.

Revision Table: Key Steps Review

Here's a quick summary of the process to find the value:

• Given \(x = 97 + 56\sqrt{3}\).
• Found \(\sqrt{x} = \sqrt{(7+4\sqrt{3})^2} = 7 + 4\sqrt{3}\).
• Found \(\sqrt[4]{x} = \sqrt{7 + 4\sqrt{3}} = \sqrt{(2+\sqrt{3})^2} = 2 + \sqrt{3}\).
• Found \(\frac{1}{\sqrt[4]{x}} = \frac{1}{2+\sqrt{3}} = 2 - \sqrt{3}\).
• Calculated \(\sqrt[4]{x} + \frac{1}{\sqrt[4]{x}} = (2+\sqrt{3}) + (2-\sqrt{3}) = 4\).

Additional Information: Simplifying Nested Radicals

The technique used here to simplify expressions like \(\sqrt{a + \sqrt{b}}\) or \(\sqrt{a + c\sqrt{d}}\) is by trying to express the term inside the square root as a perfect square \((p + q\sqrt{r})^2\). The general form for simplifying \(\sqrt{A \pm \sqrt{B}}\) is if \(A^2 - B\) is a perfect square, say \(C^2\), then \(\sqrt{A \pm \sqrt{B}} = \sqrt{\frac{A+C}{2}} \pm \sqrt{\frac{A-C}{2}}\). For expressions like \(\sqrt{a + c\sqrt{d}}\), you can rewrite it as \(\sqrt{a + \sqrt{c^2 d}}\) and then apply the formula, or directly look for \((p+q\sqrt{d})^2\). In our case, \(97 + 56\sqrt{3} = 97 + \sqrt{56^2 \times 3} = 97 + \sqrt{3136 \times 3} = 97 + \sqrt{9408}\). Here \(A=97\) and \(B=9408\). \(A^2 - B = 97^2 - 9408 = 9409 - 9408 = 1 = 1^2\). So \(C=1\).

\(\sqrt{97 + \sqrt{9408}} = \sqrt{\frac{97+1}{2}} + \sqrt{\frac{97-1}{2}} = \sqrt{\frac{98}{2}} + \sqrt{\frac{96}{2}} = \sqrt{49} + \sqrt{48} = 7 + \sqrt{16 \times 3} = 7 + 4\sqrt{3}\). This matches our earlier result, confirming the method.

Similarly, for \(\sqrt{7 + 4\sqrt{3}} = \sqrt{7 + \sqrt{16 \times 3}} = \sqrt{7 + \sqrt{48}}\). Here \(A=7\) and \(B=48\). \(A^2 - B = 7^2 - 48 = 49 - 48 = 1 = 1^2\). So \(C=1\).

\(\sqrt{7 + \sqrt{48}} = \sqrt{\frac{7+1}{2}} + \sqrt{\frac{7-1}{2}} = \sqrt{\frac{8}{2}} + \sqrt{\frac{6}{2}} = \sqrt{4} + \sqrt{3} = 2 + \sqrt{3}\). This also matches our result for \(\sqrt[4]{x}\).