What is the square root of 23 - 4√15 ?
√3 - 2√5
Finding the Square Root of \(23 - 4\sqrt{15}\)
We are asked to find the square root of the expression \(23 - 4\sqrt{15}\). Finding the square root means finding a value that, when multiplied by itself, equals the original expression \(23 - 4\sqrt{15}\).
Checking the Options by Squaring
One way to solve this type of problem, especially in a multiple-choice format, is to test each option by squaring it. We are looking for an option which, when squared, results in \(23 - 4\sqrt{15}\).
Let's consider the options provided. We will specifically look at the option \(\sqrt{3} - 2\sqrt{5}\), as it is indicated as relevant to the solution.
Calculating \((\sqrt{3} - 2\sqrt{5})^2\)
We will square the expression \(\sqrt{3} - 2\sqrt{5}\) using the algebraic identity for squaring a binomial: \((a-b)^2 = a^2 - 2ab + b^2\).
In this case, we can consider \(a = \sqrt{3}\) and \(b = 2\sqrt{5}\).
Let's calculate each term:
- The first term squared: \(a^2 = (\sqrt{3})^2 = 3\)
- Twice the product of the two terms: \(2ab = 2 \times (\sqrt{3}) \times (2\sqrt{5}) = 2 \times 2 \times \sqrt{3 \times 5} = 4\sqrt{15}\)
- The second term squared: \(b^2 = (2\sqrt{5})^2 = (2)^2 \times (\sqrt{5})^2 = 4 \times 5 = 20\)
Now, substitute these results back into the formula \((a-b)^2 = a^2 - 2ab + b^2\):
\((\sqrt{3} - 2\sqrt{5})^2 = (\sqrt{3})^2 - 2(\sqrt{3})(2\sqrt{5}) + (2\sqrt{5})^2\)
\((\sqrt{3} - 2\sqrt{5})^2 = 3 - 4\sqrt{15} + 20\)
Combine the numerical terms:
\((\sqrt{3} - 2\sqrt{5})^2 = (3 + 20) - 4\sqrt{15}\)
\((\sqrt{3} - 2\sqrt{5})^2 = 23 - 4\sqrt{15}\)
Verification of the Square Root
We have shown that squaring the expression \(\sqrt{3} - 2\sqrt{5}\) results in \(23 - 4\sqrt{15}\). This confirms that \(\sqrt{3} - 2\sqrt{5}\) is indeed a square root of \(23 - 4\sqrt{15}\).
General Method for Simplifying Nested Radicals
Expressions like \(\sqrt{a \pm \sqrt{b}}\) or \(\sqrt{a \pm c\sqrt{d}}\) can often be simplified if the expression inside the square root can be written as a perfect square, typically in the form \((\sqrt{x} \pm \sqrt{y})^2 = (x+y) \pm 2\sqrt{xy}\).
Our expression is \(\sqrt{23 - 4\sqrt{15}}\). We can rewrite \(4\sqrt{15}\) to fit the \(2\sqrt{xy}\) part of the formula. \(4\sqrt{15} = 2 \times (2\sqrt{15})\). To get the '2' inside the square root, we square it: \(2\sqrt{15} = \sqrt{2^2 \times 15} = \sqrt{4 \times 15} = \sqrt{60}\).
So, \(\sqrt{23 - 4\sqrt{15}} = \sqrt{23 - 2\sqrt{60}}\).
Now we need to find two numbers, say \(x\) and \(y\), such that their sum \(x+y = 23\) and their product \(xy = 60\). We look for pairs of factors of 60:
| Factors of 60 | Sum of Factors |
|---|---|
| 1, 60 | 61 |
| 2, 30 | 32 |
| 3, 20 | 23 |
| 4, 15 | 19 |
| 5, 12 | 17 |
| 6, 10 | 16 |
The pair (3, 20) satisfies both conditions: \(3+20=23\) and \(3 \times 20 = 60\).
So, \(23 - 2\sqrt{60}\) can be written as \(20 + 3 - 2\sqrt{20 \times 3}\). This matches the form \((\sqrt{x} - \sqrt{y})^2\) where \(x=20\) and \(y=3\).
Thus, \(\sqrt{23 - 2\sqrt{60}} = \sqrt{(\sqrt{20} - \sqrt{3})^2}\) or \(\sqrt{(\sqrt{3} - \sqrt{20})^2}\).
This simplifies to \(|\sqrt{20} - \sqrt{3}|\) or \(|\sqrt{3} - \sqrt{20}|\).
Since \(\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}\), the result is \(|2\sqrt{5} - \sqrt{3}|\) or \(|\sqrt{3} - 2\sqrt{5}|\).
The absolute value \(|z|\) is \(z\) if \(z \ge 0\) and \(-z\) if \(z < 0\). Since \(2\sqrt{5} = \sqrt{20}\) and \(\sqrt{3}\), and \(20 > 3\), \(2\sqrt{5} > \sqrt{3}\). Therefore, \(2\sqrt{5} - \sqrt{3}\) is positive, and its absolute value is \(2\sqrt{5} - \sqrt{3}\). This is the principal (non-negative) square root.
The expression \(\sqrt{3} - 2\sqrt{5}\) is negative because \(\sqrt{3} < 2\sqrt{5}\). Its absolute value is \(-( \sqrt{3} - 2\sqrt{5} ) = 2\sqrt{5} - \sqrt{3}\).
Both \(\sqrt{3} - 2\sqrt{5}\) and \(2\sqrt{5} - \sqrt{3}\) are square roots of \(23 - 4\sqrt{15}\) because squaring either expression gives \(23 - 4\sqrt{15}\). The options provided include \(\sqrt{3} - 2\sqrt{5}\), which we confirmed by squaring.
Revision Table: Simplifying Radical Expressions
| Concept | Description | Example |
|---|---|---|
| Squaring Binomials with Radicals | Using \((a \pm b)^2 = a^2 \pm 2ab + b^2\) where \(a, b\) involve radicals. | \((\sqrt{x} - \sqrt{y})^2 = x - 2\sqrt{xy} + y\) |
| Simplifying \(\sqrt{a \pm 2\sqrt{b}}\) | Find \(x, y\) s.t. \(x+y=a, xy=b\). Result is \(|\sqrt{x} \pm \sqrt{y}|\). | \(\sqrt{5 - 2\sqrt{6}} = \sqrt{(\sqrt{3}-\sqrt{2})^2} = |\sqrt{3}-\sqrt{2}| = \sqrt{3}-\sqrt{2}\) |
| Combining Like Radicals | Add or subtract terms with the same radical part. | \(3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}\) |
Additional Information: Understanding Square Roots
A square root of a number \(N\) is a number that, when squared (multiplied by itself), gives \(N\). Every positive number has two square roots, one positive and one negative.
For example, the square roots of 9 are 3 and -3, because \(3^2 = 9\) and \((-3)^2 = 9\).
The symbol \(\sqrt{}\) is typically used to denote the principal square root, which is defined as the non-negative square root. So, \(\sqrt{9} = 3\).
In this problem, \(23 - 4\sqrt{15}\) is a positive value. Its principal square root is the positive value \(2\sqrt{5} - \sqrt{3}\). However, the question asks for "the square root" and provides \(\sqrt{3} - 2\sqrt{5}\) as one of the options, which is a negative number. Since \((\sqrt{3} - 2\sqrt{5})^2 = 23 - 4\sqrt{15}\), \(\sqrt{3} - 2\sqrt{5}\) is also a square root of \(23 - 4\sqrt{15}\). When asked to select from options, the option that squares back to the original value is considered correct.
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