Solving the Nested Square Roots Problem

The problem asks us to find the value of \( \sqrt{(b-a)} \) given the equation \( \frac{\sqrt{26-7\sqrt{3} } }{\sqrt{14+5\sqrt{3} } } = \frac{b+a\sqrt{3} }{11} \), where \( b > 0 \).

To solve this, we need to simplify the terms involving nested square roots in the numerator and the denominator of the left side of the equation. We can use the formula for simplifying nested radicals: \( \sqrt{X \pm \sqrt{Y}} = \sqrt{\frac{X+\sqrt{X^2-Y}}{2}} \pm \sqrt{\frac{X-\sqrt{X^2-Y}}{2}} \), provided \( X^2-Y \) is a perfect square. For expressions like \( \sqrt{A \pm B\sqrt{C}} \), we first rewrite them as \( \sqrt{A \pm \sqrt{B^2C}} \).

Step 1: Simplify the Nested Square Roots

Simplifying the Numerator: \( \sqrt{26-7\sqrt{3}} \)

Rewrite \( \sqrt{26-7\sqrt{3}} \) as \( \sqrt{26-\sqrt{7^2 \times 3}} = \sqrt{26-\sqrt{49 \times 3}} = \sqrt{26-\sqrt{147}} \).

Using the formula with \( X=26 \) and \( Y=147 \):

\( X^2-Y = 26^2 - 147 = 676 - 147 = 529 \). Since \( \sqrt{529} = 23 \), \( X^2-Y \) is a perfect square.

So, \( \sqrt{26-\sqrt{147}} = \sqrt{\frac{26+\sqrt{529}}{2}} - \sqrt{\frac{26-\sqrt{529}}{2}} \)

\( = \sqrt{\frac{26+23}{2}} - \sqrt{\frac{26-23}{2}} = \sqrt{\frac{49}{2}} - \sqrt{\frac{3}{2}} \)

\( = \frac{\sqrt{49}}{\sqrt{2}} - \frac{\sqrt{3}}{\sqrt{2}} = \frac{7}{\sqrt{2}} - \frac{\sqrt{3}}{\sqrt{2}} = \frac{7-\sqrt{3}}{\sqrt{2}} \).

Thus, \( \sqrt{26-7\sqrt{3}} = \frac{7-\sqrt{3}}{\sqrt{2}} \).

Simplifying the Denominator: \( \sqrt{14+5\sqrt{3}} \)

Rewrite \( \sqrt{14+5\sqrt{3}} \) as \( \sqrt{14+\sqrt{5^2 \times 3}} = \sqrt{14+\sqrt{25 \times 3}} = \sqrt{14+\sqrt{75}} \).

Using the formula with \( X=14 \) and \( Y=75 \):

\( X^2-Y = 14^2 - 75 = 196 - 75 = 121 \). Since \( \sqrt{121} = 11 \), \( X^2-Y \) is a perfect square.

So, \( \sqrt{14+\sqrt{75}} = \sqrt{\frac{14+\sqrt{121}}{2}} + \sqrt{\frac{14-\sqrt{121}}{2}} \)

\( = \sqrt{\frac{14+11}{2}} + \sqrt{\frac{14-11}{2}} = \sqrt{\frac{25}{2}} + \sqrt{\frac{3}{2}} \)

\( = \frac{\sqrt{25}}{\sqrt{2}} + \frac{\sqrt{3}}{\sqrt{2}} = \frac{5}{\sqrt{2}} + \frac{\sqrt{3}}{\sqrt{2}} = \frac{5+\sqrt{3}}{\sqrt{2}} \).

Thus, \( \sqrt{14+5\sqrt{3}} = \frac{5+\sqrt{3}}{\sqrt{2}} \).

Step 2: Evaluate the Fraction and Rationalize

Now, substitute the simplified forms back into the original fraction:

\( \frac{\sqrt{26-7\sqrt{3} } }{\sqrt{14+5\sqrt{3} } } = \frac{\frac{7-\sqrt{3}}{\sqrt{2}}}{\frac{5+\sqrt{3}}{\sqrt{2}}} \)

The \( \sqrt{2} \) terms cancel out:

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

To equate this to the form \( \frac{b+a\sqrt{3}}{11} \), we need to rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is \( 5-\sqrt{3} \).

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

Expand the numerator using the FOIL method: \( (ac+bd) + (ad+bc)\sqrt{3} \)

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

\( = 35 - 7\sqrt{3} - 5\sqrt{3} + 3 = 38 - 12\sqrt{3} \)

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

\( (5+\sqrt{3})(5-\sqrt{3}) = 5^2 - (\sqrt{3})^2 = 25 - 3 = 22 \)

So the simplified fraction is:

\( \frac{38 - 12\sqrt{3}}{22} \)

We can simplify this fraction by dividing the numerator and denominator by 2:

\( \frac{38 - 12\sqrt{3}}{22} = \frac{2(19 - 6\sqrt{3})}{2 \times 11} = \frac{19 - 6\sqrt{3}}{11} \).

Step 3: Equate and Find a and b

Now, we equate the simplified left side to the right side of the given equation:

\( \frac{19 - 6\sqrt{3}}{11} = \frac{b+a\sqrt{3}}{11} \)

Since the denominators are the same, the numerators must be equal:

\( 19 - 6\sqrt{3} = b+a\sqrt{3} \)

By comparing the rational parts and the coefficients of \( \sqrt{3} \), we get:

• Rational parts: \( b = 19 \)
• Coefficients of \( \sqrt{3} \): \( a = -6 \)

The problem states that \( b > 0 \), which is true since \( b = 19 \).

Step 4: Calculate \( \sqrt{(b-a)} \)

We need to find the value of \( \sqrt{(b-a)} \). Substitute the values of \( a \) and \( b \):

\( b-a = 19 - (-6) = 19 + 6 = 25 \)

Now, calculate the square root:

\( \sqrt{(b-a)} = \sqrt{25} \)

\( \sqrt{25} = 5 \) (Since the square root symbol usually denotes the principal, non-negative root).

The value of \( \sqrt{(b-a)} \) is 5.

Revision Table: Key Concepts

This problem involved several key concepts from algebra related to manipulating radical expressions.

• Simplifying nested square roots using the formula \( \sqrt{X \pm \sqrt{Y}} \).
• Rewriting expressions in the form \( \sqrt{A \pm \sqrt{B^2C}} \).
• Rationalizing the denominator of a fraction containing radicals.
• Equating two expressions involving radicals to find unknown variables.
• Comparing rational and irrational parts of an equation like \( p+q\sqrt{r} = s+t\sqrt{r} \Rightarrow p=s, q=t \).

Additional Information: Simplifying Radicals

Simplifying radicals is a fundamental skill in algebra. Nested square roots of the form \( \sqrt{a \pm \sqrt{b}} \) or \( \sqrt{a \pm c\sqrt{d}} \) can often be simplified into the form \( \sqrt{x} \pm \sqrt{y} \) or \( p \pm q\sqrt{r} \) if certain conditions are met.

For \( \sqrt{a \pm \sqrt{b}} \), if \( a^2-b \) is a perfect square, say \( k^2 \), then \( \sqrt{a \pm \sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}} \pm \sqrt{\frac{a-\sqrt{a^2-b}}{2}} = \sqrt{\frac{a+k}{2}} \pm \sqrt{\frac{a-k}{2}} \).

For \( \sqrt{a \pm c\sqrt{d}} \), we rewrite it as \( \sqrt{a \pm \sqrt{c^2d}} \). Then apply the same formula. For the formula to work nicely, \( a^2 - c^2d \) must be a perfect square.

Alternatively, we can try to find \( x \) and \( y \) such that \( (\sqrt{x} \pm \sqrt{y})^2 = x+y \pm 2\sqrt{xy} \). If \( \sqrt{a \pm \sqrt{b}} = \sqrt{x} \pm \sqrt{y} \), then \( a \pm \sqrt{b} = x+y \pm 2\sqrt{xy} \). This implies \( a = x+y \) and \( \sqrt{b} = 2\sqrt{xy} \), so \( b = 4xy \). We need to find two numbers \( x \) and \( y \) whose sum is \( a \) and product is \( b/4 \). This approach can be easier for simpler cases but less direct for the form \( \sqrt{a \pm c\sqrt{d}} \) unless \( c \) is incorporated carefully.

Rationalizing the denominator involves multiplying the numerator and denominator by the conjugate of the denominator to eliminate the radical from the denominator. If the denominator is \( p+q\sqrt{r} \), the conjugate is \( p-q\sqrt{r} \).