if a=5-2\sqrt{6} then find the value of a^2+\frac{1}{a^2}
a = 5-2√6
a^2 = 49-20√6
1/a^2 = 49+20√6
Now just add up a^2 + 1/a^2
To find the value of \(a^2 + \frac{1}{a^2}\), let's first simplify the expression for \(a\).
Given: \(a = 5 - 2\sqrt{6}\)
We want to express \(a\) in a simple form without any radicals, so let's try squaring both sides of the equation:
\[a^2 = (5 - 2\sqrt{6})^2\]
Expanding the equation on the right side:
\[a^2 = (5 - 2\sqrt{6})(5 - 2\sqrt{6})\]
Using the FOIL method to multiply the terms:
\[a^2 = 25 - 10\sqrt{6} - 10\sqrt{6} + 24\]
Combining like terms:
\[a^2 = 49 - 20\sqrt{6}\]
Now, let's find \(\frac{1}{a^2}\):
\[\frac{1}{a^2} = \frac{1}{49 - 20\sqrt{6}}\]
To rationalize the denominator, we'll multiply both the numerator and denominator by the conjugate of the denominator:
\[\frac{1}{a^2} = \frac{1}{49 - 20\sqrt{6}} \cdot \frac{49 + 20\sqrt{6}}{49 + 20\sqrt{6}}\]
Simplifying the numerator and denominator:
\[\frac{1}{a^2} = \frac{49 + 20\sqrt{6}}{49^2 - (20\sqrt{6})^2}\]
Simplifying further:
\[\frac{1}{a^2} = \frac{49 + 20\sqrt{6}}{2401 - 720}\]
\[\frac{1}{a^2} = \frac{49 + 20\sqrt{6}}{1681}\]
\[\frac{1}{a^2} = \frac{7 + 4\sqrt{6}}{841}\]
Now, substituting the values back into the expression \(a^2 + \frac{1}{a^2}\):
\[a^2 + \frac{1}{a^2} = (49 - 20\sqrt{6}) + \frac{7 + 4\sqrt{6}}{841}\]
To simplify further, we'll combine the like terms:
\[a^2 + \frac{1}{a^2} = 49 - 20\sqrt{6} + \frac{7}{841} + \frac{4\sqrt{6}}{841}\]
Thus, the value of \(a^2 + \frac{1}{a^2}\) is \(49 - 20\sqrt{6} + \frac{7}{841} + \frac{4\sqrt{6}}{841}\).
To find the value of \(a^2 + \frac{1}{a^2}\), we need to substitute the value of \(a\) into the expression.
Given: \(a = 5 - 2\sqrt{6}\).
First, let's find the square of \(a\).
\[
a^2 = (5 - 2\sqrt{6})^2
\]
Expanding the equation using the distributive property, we have:
\[
a^2 = (5 - 2\sqrt{6}) \cdot (5 - 2\sqrt{6})
\]
Using the FOIL method, we get:
\[
a^2 = 5 \cdot 5 + 5 \cdot (-2\sqrt{6}) + (-2\sqrt{6}) \cdot 5 + (-2\sqrt{6}) \cdot (-2\sqrt{6})
\]
Simplifying further, we have:
\[
a^2 = 25 - 10\sqrt{6} - 10\sqrt{6} + 24
\]
Combining like terms, we get:
\[
a^2 = 49 - 20\sqrt{6}
\]
Now, let's find \(\frac{1}{a^2}\).
\[
\frac{1}{a^2} = \frac{1}{49 - 20\sqrt{6}}
\]
To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is \(49 + 20\sqrt{6}\).
\[
\frac{1}{a^2} = \frac{1}{49 - 20\sqrt{6}} \cdot \frac{49 + 20\sqrt{6}}{49 + 20\sqrt{6}}
\]
Multiplying the numerator and denominator, we have:
\[
\frac{1}{a^2} = \frac{49 + 20\sqrt{6}}{49^2 - (20\sqrt{6})^2}
\]
Simplifying further, we get:
\[
\frac{1}{a^2} = \frac{49 + 20\sqrt{6}}{2401 - 720}
\]
Simplifying the denominator, we have:
\[
\frac{1}{a^2} = \frac{49 + 20\sqrt{6}}{1681}
\]
Now, let's substitute the values of \(a^2\) and \(\frac{1}{a^2}\) into the expression:
\[
a^2 + \frac{1}{a^2} = (49 - 20\sqrt{6}) + \frac{49 + 20\sqrt{6}}{1681}
\]
Combining like terms, we have:
\[
a^2 + \frac{1}{a^2} = 49 - 20\sqrt{6} + \frac{49 + 20\sqrt{6}}{1681}
\]
Simplifying further, we get:
\[
a^2 + \frac{1}{a^2} = 49 - 20\sqrt{6} + \frac{49}{1681} + \frac{20\sqrt{6}}{1681}
\]
Finally, simplifying the expression, we have:
\[
a^2 + \frac{1}{a^2} = 49 - 20\sqrt{6} + \frac{49}{1681} + \frac{20\sqrt{6}}{1681}
\]