Use the discriminant to determine how many real-number solutions the equation has.
Problem 1
w2 - 2w + 2 = 0
A) 2
B) 1
C) 0
Problem 2
1 - 7a2 = -7a - 2
A) 2
B) 1
C) 0
w2 - 2w + 2 = 0
a = 1
b = -2
c = 2
b^2-4ac = 4-4(1)(2) = 4-8 = -4
sqrt (-4) is imaginary
no real solutions (although 2 complex solutions)
1 - 7a2 = -7a - 2
7a^2 -7a -3 = 0
b^2-4ac = 49 -4(7)(-3) = 133
sqrt 133 is real number >0
so two real solutions
To determine the number of real-number solutions for each equation, we can use the discriminant formula. The discriminant is found by taking the square root of b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.
Problem 1:
w^2 - 2w + 2 = 0
To find the discriminant, we need to identify the values of a, b, and c for this quadratic equation:
a = 1
b = -2
c = 2
Now, let's plug these values into the discriminant formula:
Discriminant = b^2 - 4ac
Discriminant = (-2)^2 - 4(1)(2)
Discriminant = 4 - 8
Discriminant = -4
In Problem 1, the discriminant is -4. Since the discriminant is negative, this means that the quadratic equation has no real-number solutions. Therefore, the correct answer is C) 0.
Problem 2:
1 - 7a^2 = -7a - 2
Again, let's identify the values of a, b, and c for this quadratic equation:
a = -7
b = 7
c = -2
Now, plug these values into the discriminant formula:
Discriminant = b^2 - 4ac
Discriminant = (7)^2 - 4(-7)(-2)
Discriminant = 49 - 56
Discriminant = -7
In Problem 2, the discriminant is -7. Since the discriminant is negative, this means that the quadratic equation has no real-number solutions. Therefore, the correct answer is C) 0.