What is the square root of 16 + 6√7?
3 + √7
Understanding the Problem: Finding the Square Root of a Compound Surd
The question asks us to find the square root of the expression \(16 + 6\sqrt{7}\). This type of expression, which involves a rational number added to a multiple of a square root of an irrational number, is often called a compound surd. Our goal is to simplify \(\sqrt{16 + 6\sqrt{7}}\) into a simpler form, typically \(a + \sqrt{b}\) or \(a - \sqrt{b}\).
We can approach this problem by assuming that the expression inside the square root is a perfect square of the form \((x + \sqrt{y})^2\) or \((x - \sqrt{y})^2\). Let's assume the square root is of the form \(x + \sqrt{y}\), where \(x\) is an integer and \(y\) is an integer.
Step-by-Step Solution to Finding the Square Root
We are trying to find \(\sqrt{16 + 6\sqrt{7}}\). Let's assume:
\(\sqrt{16 + 6\sqrt{7}} = x + \sqrt{y}\)
Squaring both sides of the equation, we get:
\(( \sqrt{16 + 6\sqrt{7}} )^2 = (x + \sqrt{y})^2\)
\(16 + 6\sqrt{7} = x^2 + (\sqrt{y})^2 + 2 \cdot x \cdot \sqrt{y}\)
\(16 + 6\sqrt{7} = x^2 + y + 2x\sqrt{y}\)
Now, we can equate the rational parts and the irrational parts on both sides of the equation.
The rational part on the left is \(16\). The rational part on the right is \(x^2 + y\). So, we have:
\(x^2 + y = 16 \quad \cdots (1)\)
The irrational part on the left is \(6\sqrt{7}\). The irrational part on the right is \(2x\sqrt{y}\). So, we have:
\(6\sqrt{7} = 2x\sqrt{y}\)
We can simplify the second equation:
\(3\sqrt{7} = x\sqrt{y}\)
To make it easier to compare, we can square both sides of this equation:
\((3\sqrt{7})^2 = (x\sqrt{y})^2\)
\(3^2 \cdot (\sqrt{7})^2 = x^2 \cdot (\sqrt{y})^2\)
\(9 \cdot 7 = x^2 y\)
\(63 = x^2 y \quad \cdots (2)\)
Now we have a system of two equations with two variables, \(x\) and \(y\):
- \(x^2 + y = 16\)
- \(x^2 y = 63\)
From the second equation, we can express \(y\) in terms of \(x\) (assuming \(x \neq 0\)): \(y = \frac{63}{x^2}\).
Substitute this expression for \(y\) into the first equation:
\(x^2 + \frac{63}{x^2} = 16\)
Multiply the entire equation by \(x^2\) to eliminate the denominator:
\(x^2 \cdot x^2 + x^2 \cdot \frac{63}{x^2} = 16 \cdot x^2\)
\(x^4 + 63 = 16x^2\)
Rearrange the terms to form a quadratic equation in terms of \(x^2\):
\(x^4 - 16x^2 + 63 = 0\)
Let \(z = x^2\). The equation becomes a quadratic equation in \(z\):
\(z^2 - 16z + 63 = 0\)
We can solve this quadratic equation by factoring. We are looking for two numbers that multiply to \(63\) and add up to \(-16\). These numbers are \(-7\) and \(-9\).
So, the factored equation is:
\((z - 7)(z - 9) = 0\)
This gives two possible values for \(z\):
- \(z - 7 = 0 \implies z = 7\)
- \(z - 9 = 0 \implies z = 9\)
Since \(z = x^2\), we have:
- \(x^2 = 7 \implies x = \pm \sqrt{7}\)
- \(x^2 = 9 \implies x = \pm 3\)
Recall that we assumed \(x\) is an integer. Therefore, \(x\) must be either \(3\) or \(-3\).
Now, let's find the corresponding values for \(y\) using \(y = \frac{63}{x^2}\).
- If \(x = 3\), then \(x^2 = 9\). \(y = \frac{63}{9} = 7\).
- If \(x = -3\), then \(x^2 = 9\). \(y = \frac{63}{9} = 7\).
We assumed \(\sqrt{16 + 6\sqrt{7}} = x + \sqrt{y}\). Since the left side, \(\sqrt{16 + 6\sqrt{7}}\), is a positive value (square roots are conventionally positive), the right side \(x + \sqrt{y}\) must also be positive. If \(x=3\) and \(y=7\), then \(x+\sqrt{y} = 3 + \sqrt{7}\), which is positive. If \(x=-3\) and \(y=7\), then \(x+\sqrt{y} = -3 + \sqrt{7}\). Since \(\sqrt{7}\) is approximately \(2.65\), \(-3 + \sqrt{7}\) is negative. So, we must choose the positive value for \(x\), which is \(x=3\).
Using \(x=3\) and \(y=7\), the square root is \(x + \sqrt{y} = 3 + \sqrt{7}\).
Let's verify this by squaring \((3 + \sqrt{7})\):
\((3 + \sqrt{7})^2 = 3^2 + (\sqrt{7})^2 + 2 \cdot 3 \cdot \sqrt{7}\)
\(= 9 + 7 + 6\sqrt{7}\)
\(= 16 + 6\sqrt{7}\)
This matches the expression inside the square root, so our result is correct.
Matching with the Options
Let's compare our result, \(3 + \sqrt{7}\), with the given options:
- Option 1: \(4 + \sqrt{7}\)
- Option 2: \(4 - \sqrt{7}\)
- Option 3: \(3 + \sqrt{7}\)
- Option 4: \(3 - \sqrt{7}\)
Our calculated square root is \(3 + \sqrt{7}\), which matches Option 3.
Final Answer Derivation
By assuming the square root of \(16 + 6\sqrt{7}\) is in the form \(x + \sqrt{y}\) and squaring both sides, we derived a system of equations for \(x\) and \(y\). Solving these equations yielded \(x=3\) and \(y=7\) (considering \(x\) must be positive for the principal square root). Thus, the square root is \(3 + \sqrt{7}\).
Revision Table: Square Root of Compound Surds
| Concept | Description | Example |
|---|---|---|
| Compound Surd | An expression of the form \(a \pm \sqrt{b}\) or \(a \pm c\sqrt{b}\), where \(a, b, c\) are rational numbers and \(\sqrt{b}\) is irrational. | \(5 + \sqrt{3}\), \(2 - 4\sqrt{5}\) |
| Square Root of Compound Surd | Finding \(\sqrt{a \pm \sqrt{b}}\) or \(\sqrt{a \pm c\sqrt{b}}\). Often results in the form \(\sqrt{x} \pm \sqrt{y}\) or \(p \pm \sqrt{q}\). | \(\sqrt{7 + 2\sqrt{10}} = \sqrt{(\sqrt{5} + \sqrt{2})^2} = \sqrt{5} + \sqrt{2}\) |
| Method Used | Assume \(\sqrt{A \pm \sqrt{B}} = \sqrt{x} \pm \sqrt{y}\) and square both sides. Or assume \(\sqrt{a \pm c\sqrt{b}} = p \pm \sqrt{q}\) and square both sides. Equate rational and irrational parts. | Used \(16 + 6\sqrt{7} = (x + \sqrt{y})^2 = x^2+y+2x\sqrt{y}\) to find \(x=3, y=7\). |
Additional Information: Simplifying Square Roots
The method used here is a common technique for simplifying the square root of compound surds. The general formula for \(\sqrt{a + \sqrt{b}}\) can be derived using this method. If we assume \(\sqrt{a + \sqrt{b}} = \sqrt{x} + \sqrt{y}\), squaring both sides gives \(a + \sqrt{b} = x + y + 2\sqrt{xy}\). Equating rational and irrational parts, we get \(a = x + y\) and \(\sqrt{b} = 2\sqrt{xy} \implies b = 4xy\).
From \(a = x + y\) and \(b = 4xy\), we can find \(x\) and \(y\). Consider \((x - y)^2 = (x + y)^2 - 4xy = a^2 - b\). So, \(x - y = \sqrt{a^2 - b}\). We have two equations:
- \(x + y = a\)
- \(x - y = \sqrt{a^2 - b}\)
Adding the two equations: \(2x = a + \sqrt{a^2 - b} \implies x = \frac{a + \sqrt{a^2 - b}}{2}\).
Subtracting the second equation from the first: \(2y = a - \sqrt{a^2 - b} \implies y = \frac{a - \sqrt{a^2 - b}}{2}\).
Therefore, \(\sqrt{a + \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^2 - b}}{2}} + \sqrt{\frac{a - \sqrt{a^2 - b}}{2}}\).
For the given problem, \(16 + 6\sqrt{7}\) is not directly in the form \(a + \sqrt{b}\). We have \(16 + \sqrt{36 \cdot 7} = 16 + \sqrt{252}\). So, \(a=16\) and \(b=252\).
Let's use the formula:
\(\sqrt{16 + \sqrt{252}} = \sqrt{\frac{16 + \sqrt{16^2 - 252}}{2}} + \sqrt{\frac{16 - \sqrt{16^2 - 252}}{2}}\)
\(= \sqrt{\frac{16 + \sqrt{256 - 252}}{2}} + \sqrt{\frac{16 - \sqrt{256 - 252}}{2}}\)
\(= \sqrt{\frac{16 + \sqrt{4}}{2}} + \sqrt{\frac{16 - \sqrt{4}}{2}}\)
\(= \sqrt{\frac{16 + 2}{2}} + \sqrt{\frac{16 - 2}{2}}\)
\(= \sqrt{\frac{18}{2}} + \sqrt{\frac{14}{2}}\)
\(= \sqrt{9} + \sqrt{7}\)
\(= 3 + \sqrt{7}\)
This confirms our earlier result obtained by equating coefficients, demonstrating the consistency of the methods for simplifying square roots of compound surds.
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