Answer:
Step-by-step explanation:
The discriminant is used to determine the number and nature of the zeros of a quadratic. If the discriminant is positive and a perfect square, there are 2 rational zeros; if the discriminant is positive and not a perfect square, there are 2 rational complex zeros; if the discriminant is 0, there is 1 rational root; if the discriminant is negative, there are no real roots.
The roots/solutions/zeros of a quadratic are where the graph goes through the x axis. Those are the real zeros, even if they don't fall exactly on a number like 1 or 2 or 3; they can fall on 1.32, 4.35, etc. They are still real. If the graph doesn't go through the x-axis at all, the zeros are imaginary because the discriminant was negative and you can't take the square root of a negative number. As you can see on our graph, the parabola never goes through the x-axis. Therefore, the zeros are imaginary because the discriminant was negative. Choice C. Get familiar with your discriminants and the nature of quadratic solutions. Your life will be much easier!
It's actually easier than it seems. if the tax is paid in equal payments every month and the total throughout a year (12 months) is $2,820, you have to find out how much they pay every month
so divide the total by the number of months
2,820/12=235
they pay $235 in taxes every month
to find out the total of the taxes and the house payment, just add the monthly tax and the monthly house payment.
235+752
and your answer is the sum
Answer:
x = -2,-3
Step-by-step explanation:
Solve the equation by factorizing.
To factorize this equation, you have to find the two numbers that adds up to 5 and multiplies to 6.
This one is obvious, but for others you may use trial and error.
After finding the 2 numbers: 2 and 3.
Present them in this format:
(x+3) ( x+2) = 0
You may check if this is correct using the distribution method.
After this,
you have to solve it individually.
for example:
x+3 = 0
x = -3
x+2 = 0
x = -2.
Thus, x has 2 different values: -3, -2.
You may check this by substituting x with either of these values.
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