Certainly! Let's tackle the problem step by step to find the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] for the given expression [tex]\(\sqrt{6-4\sqrt{2}} = \sqrt{a} - \sqrt{b}\)[/tex].
1. Start with the given equation:
[tex]\[
\sqrt{6-4\sqrt{2}} = \sqrt{a} - \sqrt{b}
\][/tex]
2. Square both sides to eliminate the square roots:
[tex]\[
( \sqrt{6-4\sqrt{2}} )^2 = ( \sqrt{a} - \sqrt{b} )^2
\][/tex]
[tex]\[
6 - 4\sqrt{2} = a + b - 2\sqrt{ab}
\][/tex]
3. Separate the rational and irrational parts. Equate the rational parts and the irrational parts separately:
- The rational part: [tex]\[
6 = a + b
\][/tex]
- The irrational part: [tex]\[
-4\sqrt{2} = -2\sqrt{ab}
\][/tex]
4. Simplify the irrational part equation:
[tex]\[
-4\sqrt{2} = -2\sqrt{ab}
\][/tex]
[tex]\[
4\sqrt{2} = 2\sqrt{ab}
\][/tex]
[tex]\[
2\sqrt{2} = \sqrt{ab}
\][/tex]
[tex]\[
(2\sqrt{2})^2 = ab
\][/tex]
[tex]\[
8 = ab
\][/tex]
5. Now we have a system of equations:
[tex]\[
\begin{cases}
a + b = 6 \\
ab = 8
\end{cases}
\][/tex]
6. Solve the system of equations:
To find [tex]\( a \)[/tex] and [tex]\( b \)[/tex], let's solve these two equations simultaneously.
- Let’s consider the quadratic equation that satisfies these conditions: [tex]\[
t^2 - (a+b)t + ab = 0
\][/tex]
Plugging in [tex]\( a+b = 6 \)[/tex] and [tex]\( ab = 8 \)[/tex]:
[tex]\[
t^2 - 6t + 8 = 0
\][/tex]
- Solve this quadratic equation using the quadratic formula [tex]\( t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \)[/tex]:
[tex]\[
t = \frac{6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}
\][/tex]
[tex]\[
t = \frac{6 \pm \sqrt{36 - 32}}{2}
\][/tex]
[tex]\[
t = \frac{6 \pm \sqrt{4}}{2}
\][/tex]
[tex]\[
t = \frac{6 \pm 2}{2}
\][/tex]
So, we get two solutions: [tex]\[
t = \frac{6 + 2}{2} = 4 \quad \text{and} \quad t = \frac{6 - 2}{2} = 2
\][/tex]
7. Hence, the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are:
[tex]\[
a = 4, \quad b = 2 \quad \text{or} \quad a = 2, \quad b = 4
\][/tex]
So, the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are [tex]\( 2 \)[/tex] and [tex]\( 4 \)[/tex].
