w2 - 2w + 2 = 0
Use the discriminant to determine how many real-number solutions the equation has
A.2
B.1
C.0
The discriminant is b^2-4ac. For the equation you posted, a=1 (the coefficient on w^2), b=-2, and c=2. Plug those numbers in. If the number you get is positive, there are 2 real number solutions. If you get zero, there is 1 solution. If you get a negative number, there are no solutions. Post your answer if you want it checked.
Actually, if you get a negative discriminant, you get no real solutions, but insteady, you get complex (in the imaginary domain) solutions, that in general, are valid.
eg. x^2+0x+1=0
the discriminate is sqrt(-4)
and in fact, then the solution is
x= +- i
Is the answer 0 not sure if this is correct.
Show me your work. You have to plug in the numbers.
To determine the number of real-number solutions of the equation "w2 - 2w + 2 = 0" using the discriminant, you need to calculate the discriminant first. The discriminant is a value derived from the coefficients of a quadratic equation and is given by the formula:
Discriminant (D) = b^2 - 4ac
Where:
- a is the coefficient of the squared term (w^2 in this case),
- b is the coefficient of the linear term (-2w in this case), and
- c is the constant term (2 in this case).
Now let's substitute the values into the formula:
D = (-2)^2 - 4(1)(2)
= 4 - 8
= -4
After calculating the discriminant, we find that D = -4.
To determine the number of real-number solutions, you need to consider the value of the discriminant:
1. If the discriminant is positive (D > 0), then the quadratic equation has two distinct real-number solutions.
2. If the discriminant is zero (D = 0), then the quadratic equation has one real-number solution.
3. If the discriminant is negative (D < 0), then the quadratic equation has no real-number solutions.
In this case, the discriminant D = -4, which is negative (D < 0). Therefore, the equation "w2 - 2w + 2 = 0" has no real-number solutions.
So the correct answer is C.0