We are given:

1. $x + y = 8$
2. $\log_2(xy) = 2$

Step 1: Simplify the second equation

From $\log_2(xy) = 2$, we rewrite this in exponential form:

$xy = 2^2 = 4$

So, $xy = 4$.

Step 2: Solve the system of equations

We now have two equations:

1. $x + y = 8$
2. $xy = 4$

Let $x$ and $y$ be the roots of a quadratic equation. A quadratic equation with roots $x$ and $y$ can be written as:

$t^2 - (x + y)t + xy = 0$

Substitute $x + y = 8$ and $xy = 4$:

$t^2 - 8t + 4 = 0$

Step 3: Solve the quadratic equation

The quadratic equation is:

$t^2 - 8t + 4 = 0$

Use the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = -8$, and $c = 4$. Substitute these values:

$t = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(4)}}{2(1)}$ $t = \frac{8 \pm \sqrt{64 - 16}}{2}$ $t = \frac{8 \pm \sqrt{48}}{2}$ $t = \frac{8 \pm 4\sqrt{3}}{2}$ $t = 4 \pm 2\sqrt{3}$

Step 4: Assign $x$ and $y$

The roots are $t_1 = 4 + 2\sqrt{3}$ and $t_2 = 4 - 2\sqrt{3}$. Since $x + y = 8$, the values of $x$ and $y$ are:

$x = 4 + 2\sqrt{3}, \quad y = 4 - 2\sqrt{3}$

or vice versa.

Thus, $y = 4 - 2\sqrt{3}$ (assuming $y < x$).

Final Answer:

$y = 4 - 2\sqrt{3}$

Would you like me to clarify any step further?

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Related Questions:

1. How do you simplify quadratic equations further when roots involve square roots?
2. Can $\log_2(xy) = 2$ be solved for $x$ and $y$ without using the quadratic method?
3. What if the logarithm had a different base? How would it change the process?
4. How does the discriminant of a quadratic equation determine the nature of its roots?
5. Can $x$ and $y$ ever be equal in such cases? Why or why not?

Tip: Always verify solutions by substituting back into the original equations!