To solve the expression [tex]\(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}\)[/tex], let's denote it by [tex]\( x \)[/tex]. So, we have:
[tex]\[ x = \sqrt{6+\sqrt{11}} - \sqrt{6-\sqrt{11}} \][/tex]
First, we square both sides of the equation:
[tex]\[ x^2 = \left( \sqrt{6+\sqrt{11}} - \sqrt{6-\sqrt{11}} \right)^2 \][/tex]
Now, let's use the binomial theorem for squaring two terms:
[tex]\[ x^2 = \left( \sqrt{6+\sqrt{11}} \right)^2 - 2 \cdot \sqrt{6+\sqrt{11}} \cdot \sqrt{6-\sqrt{11}} + \left( \sqrt{6-\sqrt{11}} \right)^2 \][/tex]
Simplifying the squares of the square roots, we get:
[tex]\[ x^2 = (6+\sqrt{11}) + (6-\sqrt{11}) - 2 \cdot \sqrt{(6+\sqrt{11})(6-\sqrt{11})} \][/tex]
Next, we simplify the terms inside the square root. The expression [tex]\((6+\sqrt{11})(6-\sqrt{11})\)[/tex] is a difference of squares:
[tex]\[ (6+\sqrt{11})(6-\sqrt{11}) = 6^2 - (\sqrt{11})^2 \][/tex]
Calculating these values:
[tex]\[ = 36 - 11 \][/tex]
[tex]\[ = 25 \][/tex]
So, the expression simplifies to:
[tex]\[ x^2 = 6 + \sqrt{11} + 6 - \sqrt{11} - 2\sqrt{25} \][/tex]
The terms [tex]\(\sqrt{11}\)[/tex] and [tex]\(-\sqrt{11}\)[/tex] cancel each other out:
[tex]\[ x^2 = 6 + 6 - 2 \cdot 5 \][/tex]
[tex]\[ x^2 = 12 - 10 \][/tex]
[tex]\[ x^2 = 2 \][/tex]
Finally, we take the square root of both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \sqrt{2} \][/tex]
Thus, the solution to the expression [tex]\(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}\)[/tex] is:
[tex]\[ \sqrt{2} \approx 1.4142135623730951 \][/tex]
So, [tex]\(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}} = \sqrt{2} \approx 1.4142135623730951\)[/tex].
