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