Questions: Question 5 Solve the equation. y^2 - 6y + 5 = 0 y = Question 6 Solve the following equation. x^2 - 4x = 0 x = Question 7 Compute the product. (p-2)^2 =

Solution Steps

1. Question 5: The equation $y^2 - 6y + 5 = 0$ is a quadratic equation. We can solve it using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -6$, and $c = 5$.

2. Question 6: The equation $x^2 - 4x = 0$ can be solved by factoring. We can factor out an $x$ to get $x(x - 4) = 0$, which gives the solutions $x = 0$ and $x = 4$.

3. Question 7: To compute the product $(p-2)^2$, we can expand it using the formula for the square of a binomial: $(a-b)^2 = a^2 - 2ab + b^2$.

To solve the quadratic equation $y^2 - 6y + 5 = 0$, we calculate the discriminant:

$$\text{discriminant} = b^2 - 4ac = (-6)^2 - 4 \cdot 1 \cdot 5 = 36 - 20 = 16$$

Since the discriminant is positive, there are two real solutions. We apply the quadratic formula:

$$y = \frac{-b \pm \sqrt{\text{discriminant}}}{2a} = \frac{6 \pm \sqrt{16}}{2 \cdot 1} = \frac{6 \pm 4}{2}$$

Calculating the two solutions:

$$y_1 = \frac{6 + 4}{2} = \frac{10}{2} = 5$$ $$y_2 = \frac{6 - 4}{2} = \frac{2}{2} = 1$$

Thus, the solutions for $y$ are $y = 5$ and $y = 1$.

We can factor the equation $x^2 - 4x = 0$:

$$x(x - 4) = 0$$

Setting each factor to zero gives us the solutions:

$$x_1 = 0 \quad \text{and} \quad x_2 = 4$$

Thus, the solutions for $x$ are $x = 0$ and $x = 4$.

To compute the product $(p-2)^2$, we substitute $p = 2$:

$$(p - 2)^2 = (2 - 2)^2 = 0^2 = 0$$

Final Answer