Question 5: The equation y2−6y+5=0 is a quadratic equation. We can solve it using the quadratic formula y=2a−b±b2−4ac, where a=1, b=−6, and c=5.
Question 6: The equation x2−4x=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.
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=a2−2ab+b2.
To solve the quadratic equation y2−6y+5=0, we calculate the discriminant:
discriminant=b2−4ac=(−6)2−4⋅1⋅5=36−20=16
Since the discriminant is positive, there are two real solutions. We apply the quadratic formula:
y=2a−b±discriminant=2⋅16±16=26±4
Calculating the two solutions:
y1=26+4=210=5
y2=26−4=22=1
Thus, the solutions for y are y=5 and y=1.
We can factor the equation x2−4x=0:
x(x−4)=0
Setting each factor to zero gives us the solutions:
x1=0andx2=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=02=0
The solutions are:
- For y: y=5 and y=1
- For x: x=0 and x=4
- For the product: 0