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 =

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 =
Transcript text: Question 5 Solve the equation. \[ \begin{array}{l} y^{2}-6 y+5=0 \\ y=\square \end{array} \] Question 6 Solve the following equation. \[ \begin{array}{l} x^{2}-4 x=0 \\ x=\square \end{array} \] Question 7 Compute the product. \[ (p-2)^{2}= \] $\square$
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Solution

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Solution Steps

Solution Approach
  1. Question 5: The equation y26y+5=0 y^2 - 6y + 5 = 0 is a quadratic equation. We can solve it using the quadratic formula y=b±b24ac2a y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , where a=1 a = 1 , b=6 b = -6 , and c=5 c = 5 .

  2. Question 6: The equation x24x=0 x^2 - 4x = 0 can be solved by factoring. We can factor out an x x to get x(x4)=0 x(x - 4) = 0 , which gives the solutions x=0 x = 0 and x=4 x = 4 .

  3. Question 7: To compute the product (p2)2(p-2)^2, we can expand it using the formula for the square of a binomial: (ab)2=a22ab+b2(a-b)^2 = a^2 - 2ab + b^2.

Step 1: Solve the Quadratic Equation y26y+5=0 y^2 - 6y + 5 = 0

To solve the quadratic equation y26y+5=0 y^2 - 6y + 5 = 0 , we calculate the discriminant:

discriminant=b24ac=(6)2415=3620=16 \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=b±discriminant2a=6±1621=6±42 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:

y1=6+42=102=5 y_1 = \frac{6 + 4}{2} = \frac{10}{2} = 5 y2=642=22=1 y_2 = \frac{6 - 4}{2} = \frac{2}{2} = 1

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

Step 2: Solve the Equation x24x=0 x^2 - 4x = 0

We can factor the equation x24x=0 x^2 - 4x = 0 :

x(x4)=0 x(x - 4) = 0

Setting each factor to zero gives us the solutions:

x1=0andx2=4 x_1 = 0 \quad \text{and} \quad x_2 = 4

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

Step 3: Compute the Product (p2)2 (p-2)^2

To compute the product (p2)2 (p-2)^2 , we substitute p=2 p = 2 :

(p2)2=(22)2=02=0 (p - 2)^2 = (2 - 2)^2 = 0^2 = 0

Final Answer

The solutions are:

  • For y y : y=5 \boxed{y = 5} and y=1 \boxed{y = 1}
  • For x x : x=0 \boxed{x = 0} and x=4 \boxed{x = 4}
  • For the product: 0 \boxed{0}
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