The value of \(\frac{7+3\sqrt5}{3+\sqrt5}-\frac{7-3\sqrt5}{3-\sqrt{5}}\) lies between:
2 and 2.5
Simplifying Radical Expressions: Step-by-Step Solution
The question asks us to find the value of a given expression involving square roots and determine which range it falls into. The expression is: \(\frac{7+3\sqrt5}{3+\sqrt5}-\frac{7-3\sqrt5}{3-\sqrt{5}}\).
To simplify this expression, we will rationalize the denominators of each fraction.
Rationalizing the First Term
The first term is \(\frac{7+3\sqrt5}{3+\sqrt5}\). To rationalize the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator, which is \(3-\sqrt5\).
Expression for the first term:
\[ \frac{7+3\sqrt5}{3+\sqrt5} = \frac{(7+3\sqrt5)(3-\sqrt5)}{(3+\sqrt5)(3-\sqrt5)} \]
Let's calculate the numerator and the denominator separately.
Numerator:
\[ (7+3\sqrt5)(3-\sqrt5) = 7(3) - 7(\sqrt5) + (3\sqrt5)(3) - (3\sqrt5)(\sqrt5) \]
\[ = 21 - 7\sqrt5 + 9\sqrt5 - 3(\sqrt5 \times \sqrt5) \]
\[ = 21 - 7\sqrt5 + 9\sqrt5 - 3(5) \]
\[ = 21 + 2\sqrt5 - 15 \]
\[ = 6 + 2\sqrt5 \]
Denominator:
\[ (3+\sqrt5)(3-\sqrt5) \]
Using the difference of squares formula, \((a+b)(a-b) = a^2 - b^2\):
\[ (3+\sqrt5)(3-\sqrt5) = 3^2 - (\sqrt5)^2 \]
\[ = 9 - 5 \]
\[ = 4 \]
So, the first term simplifies to:
\[ \frac{6 + 2\sqrt5}{4} = \frac{2(3 + \sqrt5)}{4} = \frac{3 + \sqrt5}{2} \]
Rationalizing the Second Term
The second term is \(\frac{7-3\sqrt5}{3-\sqrt{5}}\). To rationalize its denominator, we multiply both the numerator and the denominator by the conjugate of the denominator, which is \(3+\sqrt5\).
Expression for the second term:
\[ \frac{7-3\sqrt5}{3-\sqrt{5}} = \frac{(7-3\sqrt5)(3+\sqrt5)}{(3-\sqrt{5})(3+\sqrt5)} \]
Let's calculate the numerator and the denominator separately.
Numerator:
\[ (7-3\sqrt5)(3+\sqrt5) = 7(3) + 7(\sqrt5) - (3\sqrt5)(3) - (3\sqrt5)(\sqrt5) \]
\[ = 21 + 7\sqrt5 - 9\sqrt5 - 3(\sqrt5 \times \sqrt5) \]
\[ = 21 + 7\sqrt5 - 9\sqrt5 - 3(5) \]
\[ = 21 - 2\sqrt5 - 15 \]
\[ = 6 - 2\sqrt5 \]
Denominator:
\[ (3-\sqrt5)(3+\sqrt5) \]
Using the difference of squares formula, \((a-b)(a+b) = a^2 - b^2\):
\[ (3-\sqrt5)(3+\sqrt5) = 3^2 - (\sqrt5)^2 \]
\[ = 9 - 5 \]
\[ = 4 \]
So, the second term simplifies to:
\[ \frac{6 - 2\sqrt5}{4} = \frac{2(3 - \sqrt5)}{4} = \frac{3 - \sqrt5}{2} \]
Subtracting the Simplified Terms
Now we subtract the simplified second term from the simplified first term:
\[ \left(\frac{3 + \sqrt5}{2}\right) - \left(\frac{3 - \sqrt5}{2}\right) \]
Since the denominators are the same, we can combine the numerators:
\[ = \frac{(3 + \sqrt5) - (3 - \sqrt5)}{2} \]
\[ = \frac{3 + \sqrt5 - 3 + \sqrt5}{2} \]
\[ = \frac{2\sqrt5}{2} \]
\[ = \sqrt5 \]
Determining the Range
The value of the expression is \(\sqrt5\). We need to find which range this value lies in.
We know that \(2^2 = 4\) and \(3^2 = 9\). Since \(4 < 5 < 9\), we have \(\sqrt{4} < \sqrt{5} < \sqrt{9}\), which means \(2 < \sqrt{5} < 3\).
To be more precise, let's consider the given options:
- 1.5 and 2
- 3 and 3.5
- 2.5 and 3
- 2 and 2.5
We know \(\sqrt{5}\) is between 2 and 3. Let's consider the value of \(\sqrt{5}\) more accurately. We know \(2.2^2 = 4.84\) and \(2.3^2 = 5.29\). Since \(4.84 < 5 < 5.29\), we have \(2.2 < \sqrt{5} < 2.3\).
The value of \(\sqrt5\) is approximately 2.236.
Now let's check which of the given ranges contains 2.236:
- 1.5 and 2: Does not contain 2.236.
- 3 and 3.5: Does not contain 2.236.
- 2.5 and 3: Does not contain 2.236 (2.236 is less than 2.5).
- 2 and 2.5: Contains 2.236 (since 2 < 2.236 < 2.5).
Therefore, the value of the expression lies between 2 and 2.5.
| Term | Original Form | Rationalized Form |
|---|---|---|
| First Term | \(\frac{7+3\sqrt5}{3+\sqrt5}\) | \(\frac{3+\sqrt5}{2}\) |
| Second Term | \(\frac{7-3\sqrt5}{3-\sqrt{5}}\) | \(\frac{3-\sqrt5}{2}\) |
| Approximation | Value |
|---|---|
| \(\sqrt{4}\) | 2 |
| \(\sqrt{5}\) | \(\approx 2.236\) |
| \(\sqrt{9}\) | 3 |
Conclusion
The simplified value of the expression \(\frac{7+3\sqrt5}{3+\sqrt5}-\frac{7-3\sqrt5}{3-\sqrt{5}}\) is \(\sqrt5\), which is approximately 2.236. This value lies between 2 and 2.5.
Revision Table: Radical Expression Simplification
| Concept | Description | Formula/Example |
|---|---|---|
| Rationalizing Denominator | Eliminating radicals from the denominator of a fraction by multiplying by a suitable expression (often the conjugate). | \(\frac{1}{\sqrt{a}} = \frac{1 \times \sqrt{a}}{\sqrt{a} \times \sqrt{a}} = \frac{\sqrt{a}}{a}\) \(\frac{1}{a+\sqrt{b}} = \frac{1 \times (a-\sqrt{b})}{(a+\sqrt{b})(a-\sqrt{b})} = \frac{a-\sqrt{b}}{a^2-b}\) |
| Conjugate of a Binomial Surd | For a binomial surd like \(a+\sqrt{b}\), its conjugate is \(a-\sqrt{b}\). For \(a-\sqrt{b}\), its conjugate is \(a+\sqrt{b}\). | Conjugate of \(3+\sqrt5\) is \(3-\sqrt5\). Conjugate of \(7-2\sqrt{3}\) is \(7+2\sqrt{3}\). |
| Difference of Squares | A fundamental algebraic identity used in rationalizing denominators. | \((a+b)(a-b) = a^2 - b^2\) |
| Approximating Square Roots | Estimating the value of a square root by comparing it to known squares. | \(\sqrt{5}\) is between \(\sqrt{4}=2\) and \(\sqrt{9}=3\). |
Additional Information: Properties of Radicals
Radicals (or roots) are mathematical expressions that use the radical symbol (\(\sqrt{\hspace{0.2em}\cdot\hspace{0.2em}}\)). They are the inverse operation of exponentiation.
- Product Property: \(\sqrt{ab} = \sqrt{a} \times \sqrt{b}\) (for non-negative a and b)
- Quotient Property: \(\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\) (for non-negative a and positive b)
- Adding/Subtracting Radicals: Radicals can be added or subtracted only if they have the same radicand (the number inside the square root symbol). For example, \(2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}\), but \(2\sqrt{3} + 5\sqrt{2}\) cannot be simplified further.
- Simplifying Radicals: We can simplify a radical by factoring out perfect squares from the radicand. For example, \(\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}\).
Rationalization of the denominator is a standard technique to simplify expressions, especially before performing addition or subtraction of fractions involving radicals, as it often leads to a common, non-radical denominator.
Important Questions from Simplification
Simplify the following expression.
\(\left(\frac{7}{16} \div \frac{1}{2}\:of\: \frac{1}{5}\right)\times \frac{4}{5}-\frac{1}{3}\times\frac{5}{8}\div \frac{1}{2}+\frac{3}{4}\)
The value of \(\left( {2\frac{6}{7}of4\frac{1}{5} \div \frac{2}{3}} \right) \times 5\frac{1}{9} \div \left( {\frac{3}{4} \times 2\frac{2}{3}of\frac{1}{2} \div \frac{1}{4}} \right)\) is:
The value of \(\left[ {\frac{4}{7}\rm \;of\;2\frac{4}{5} \times 1\frac{2}{3} - \left( {3\frac{1}{2} - 2\frac{1}{6}} \right)} \right] \div \left( {3\frac{1}{5} \div 4\frac{1}{2}\;\rm of\;\;5\frac{1}{3}} \right)\) is:
The value of \(\frac{{0.0203 \times 2.92}}{{0.7 \times 0.0365 \times 2.9}} \div \frac{{{{\left( {12.12} \right)}^2} - {{\left( {8.12} \right)}^2}}}{{{{\left( {0.25} \right)}^2} + \left( {0.25} \right)\left( {19.99} \right)}}\) is:
The value of 4 ÷ 12 of [3 ÷ 4 of {(4 - 2) × 6 ÷ 2}] - 2 × 6 ÷ 8 + 3 is:
