Solving Radical Equations with Nested Square Roots

The problem asks us to simplify a fraction involving nested square roots and then use the result to find the value of \(b-a\) by comparing it with a given expression \(\frac{a+b\sqrt{3}}{23}\).

The given equation is:

\(\frac{\sqrt{38-5\sqrt{3} } }{\sqrt{26+7\sqrt{3} } }= \frac{a+b\sqrt{3} }{23}\)

We need to simplify the numerator and the denominator separately.

Understanding Nested Square Roots Simplification

A common technique to simplify nested square roots of the form \(\sqrt{A \pm \sqrt{B}}\) is to look for two numbers \(x\) and \(y\) such that \(( \sqrt{x} \pm \sqrt{y} )^2 = x+y \pm 2\sqrt{xy}\). If the term inside the outer square root is \(A \pm C\sqrt{D}\), we can sometimes rewrite it as \(\sqrt{A \pm \sqrt{C^2D}}\) and then look for \(x, y\) such that \(x+y=A\) and \(xy=C^2D/4\) (if C is even, or multiply by 2/2 to make the term next to the root 2). A more general formula for \(\sqrt{A \pm \sqrt{B}}\) is \(\sqrt{\frac{A+\sqrt{A^2-B}}{2}} \pm \sqrt{\frac{A-\sqrt{A^2-B}}{2}}\), provided \(A^2-B\) is a perfect square.

Simplifying the Numerator \(\sqrt{38-5\sqrt{3}}\)

The numerator is \(\sqrt{38-5\sqrt{3}}\). We can use the formula \(\sqrt{A-\sqrt{B}}\) where \(A=38\) and \(B=(5\sqrt{3})^2 = 25 \times 3 = 75\).

First, calculate \(A^2-B = 38^2 - 75 = 1444 - 75 = 1369\).

Since \(1369 = 37^2\), \(A^2-B\) is a perfect square.

Using the formula:

\(\sqrt{38-5\sqrt{3}} = \sqrt{\frac{38+\sqrt{1369}}{2}} - \sqrt{\frac{38-\sqrt{1369}}{2}}\)

\(= \sqrt{\frac{38+37}{2}} - \sqrt{\frac{38-37}{2}}\)

\(= \sqrt{\frac{75}{2}} - \sqrt{\frac{1}{2}}\)

\(= \frac{\sqrt{75}}{\sqrt{2}} - \frac{\sqrt{1}}{\sqrt{2}}\)

\(= \frac{5\sqrt{3}}{\sqrt{2}} - \frac{1}{\sqrt{2}}\)

\(= \frac{5\sqrt{3}-1}{\sqrt{2}}\)

Simplifying the Denominator \(\sqrt{26+7\sqrt{3}}\)

The denominator is \(\sqrt{26+7\sqrt{3}}\). We use the formula \(\sqrt{A+\sqrt{B}}\) where \(A=26\) and \(B=(7\sqrt{3})^2 = 49 \times 3 = 147\).

First, calculate \(A^2-B = 26^2 - 147 = 676 - 147 = 529\).

Since \(529 = 23^2\), \(A^2-B\) is a perfect square.

Using the formula:

\(\sqrt{26+7\sqrt{3}} = \sqrt{\frac{26+\sqrt{529}}{2}} + \sqrt{\frac{26-\sqrt{529}}{2}}\)

\(= \sqrt{\frac{26+23}{2}} + \sqrt{\frac{26-23}{2}}\)

\(= \sqrt{\frac{49}{2}} + \sqrt{\frac{3}{2}}\)

\(= \frac{\sqrt{49}}{\sqrt{2}} + \frac{\sqrt{3}}{\sqrt{2}}\)

\(= \frac{7}{\sqrt{2}} + \frac{\sqrt{3}}{\sqrt{2}}\)

\(= \frac{7+\sqrt{3}}{\sqrt{2}}\)

Dividing and Rationalizing the Expression

Now we substitute the simplified numerator and denominator back into the original fraction:

\(\frac{\sqrt{38-5\sqrt{3} } }{\sqrt{26+7\sqrt{3} } } = \frac{\frac{5\sqrt{3}-1}{\sqrt{2}}}{\frac{7+\sqrt{3}}{\sqrt{2}}}\)

The \(\sqrt{2}\) terms in the denominator of both fractions cancel out:

\(= \frac{5\sqrt{3}-1}{7+\sqrt{3}}\)

To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is \(7-\sqrt{3}\).

\(= \frac{(5\sqrt{3}-1)(7-\sqrt{3})}{(7+\sqrt{3})(7-\sqrt{3})}\)

Expand the numerator using the distributive property (FOIL):

• \(5\sqrt{3} \times 7 = 35\sqrt{3}\)
• \(5\sqrt{3} \times (-\sqrt{3}) = -5 \times 3 = -15\)
• \(-1 \times 7 = -7\)
• \(-1 \times (-\sqrt{3}) = +\sqrt{3}\)

Numerator: \(35\sqrt{3} - 15 - 7 + \sqrt{3} = (35+1)\sqrt{3} + (-15-7) = 36\sqrt{3} - 22\).

Expand the denominator using the difference of squares formula \((x+y)(x-y) = x^2 - y^2\):

Denominator: \((7+\sqrt{3})(7-\sqrt{3}) = 7^2 - (\sqrt{3})^2 = 49 - 3 = 46\).

So the simplified fraction is:

\(= \frac{36\sqrt{3} - 22}{46}\)

Factor out 2 from the numerator and the denominator:

\(= \frac{2(18\sqrt{3} - 11)}{2(23)}\)

\(= \frac{18\sqrt{3} - 11}{23}\)

We can rearrange the terms in the numerator to match the form \(a+b\sqrt{3}\):

\(= \frac{-11 + 18\sqrt{3}}{23}\)

Comparing to Find a and b Values

We are given that \(\frac{\sqrt{38-5\sqrt{3} } }{\sqrt{26+7\sqrt{3} } }= \frac{a+b\sqrt{3} }{23}\).

We found that \(\frac{\sqrt{38-5\sqrt{3} } }{\sqrt{26+7\sqrt{3} } } = \frac{-11 + 18\sqrt{3}}{23}\).

Comparing \(\frac{-11 + 18\sqrt{3}}{23}\) with \(\frac{a+b\sqrt{3}}{23}\), we can equate the numerators:

\(a + b\sqrt{3} = -11 + 18\sqrt{3}\)

By comparing the rational and irrational parts, we get:

• \(a = -11\)
• \(b = 18\)

We check the condition \(b > 0\). \(18 > 0\), which is satisfied.

Calculating the Final Value (b-a)

Finally, we need to find the value of \(b-a\).

\(b - a = 18 - (-11)\)

\(= 18 + 11\)

\(= 29\)

Revision Table for Radical Problems

Additional Information on Simplifying Radicals

Simplifying expressions with radicals is a key skill in algebra. Nested radicals can often be simplified if the expression inside the outer root can be written as a perfect square. For \(\sqrt{A \pm 2\sqrt{B}}\), we look for two numbers \(x\) and \(y\) such that \(x+y=A\) and \(xy=B\). Then \(\sqrt{A \pm 2\sqrt{B}} = \sqrt{( \sqrt{x} \pm \sqrt{y} )^2} = |\sqrt{x} \pm \sqrt{y}|\). If there is a coefficient other than 2 before the inner radical, like \(\sqrt{A \pm C\sqrt{D}}\), we can try rewriting it as \(\sqrt{A \pm \sqrt{C^2D}}\) and then applying the formula for \(\sqrt{A' \pm \sqrt{B'}}\). In this problem, we successfully used the general formula approach.

Rationalization is essential when dealing with fractions involving radicals in the denominator. It makes the expression easier to work with and is considered standard form. The conjugate of \(x+y\sqrt{z}\) is \(x-y\sqrt{z}\), and their product \((x+y\sqrt{z})(x-y\sqrt{z}) = x^2 - (y\sqrt{z})^2 = x^2 - y^2z\) is a rational number (if \(x, y, z\) are rational).

Comparing the coefficients of rational and irrational parts is a valid method for solving equations involving surds, based on the fact that a rational number cannot equal a non-zero irrational number (or a combination like \(p + q\sqrt{r}\) unless \(p=0\) and \(q=0\) or \(p=s\) and \(q=t\) if \(p+q\sqrt{r} = s+t\sqrt{r}\)).