The problem involves solving the quadratic equation $x^2 - 6x + 4 = 0$ and expressing the solutions in the form $a \pm \sqrt{b}$, where $a$ and $b$ are prime numbers.

Step-by-step Solution:

1. Identify the coefficients: The given equation is: $x^2 - 6x + 4 = 0$ Here, $a = 1$, $b = -6$, and $c = 4$ (the coefficients of $x^2$, $x$, and the constant term, respectively).

2. Use the quadratic formula: The solutions for $x$ in a quadratic equation $ax^2 + bx + c = 0$ are given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Substituting the values $a = 1$, $b = -6$, and $c = 4$ into the formula: $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(4)}}{2(1)}$

3. Simplify the discriminant: The discriminant is: $(-6)^2 - 4(1)(4) = 36 - 16 = 20$ So, the solutions are: $x = \frac{6 \pm \sqrt{20}}{2}$

4. Simplify the square root: Factorize $20$ as $4 \times 5$: $\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$ Substituting this back, we get: $x = \frac{6 \pm 2\sqrt{5}}{2}$

5. Simplify the fraction: Divide both terms in the numerator by 2: $x = 3 \pm \sqrt{5}$

6. Identify $a$ and $b$: The solutions are written as $a \pm \sqrt{b}$. Comparing this with $3 \pm \sqrt{5}$, we find: $a = 3 \quad \text{and} \quad b = 5$

Final Answer:

$a = 3, \, b = 5$

Would you like a detailed explanation of any specific step?

Related Questions:

1. What are the conditions for the discriminant ($b^2 - 4ac$) to ensure real roots?
2. How do you determine if a number is prime?
3. Can quadratic equations always be solved using the quadratic formula?
4. What does it mean for solutions to be in the form $a \pm \sqrt{b}$?
5. How do the coefficients $a$, $b$, and $c$ influence the solutions of a quadratic equation?

Tip:

When solving quadratic equations, always simplify the square root fully to find the most compact form of the solution!