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kompoz [17]
3 years ago
15

Anderson uses the discriminant to correctly find the number of real solutions of the quadratic equation x2 + 4x + 8 = 0. Which e

xplanation could Anderson provide?
Mathematics
1 answer:
astra-53 [7]3 years ago
8 0

Answer:

As the result of our discriminant is negative, this quadratic equation will have two complex solutions.

Step-by-step explanation:

Recall that the formula for a discriminant is b^2-4ac

From the given quadratic equation x^2+4x+8=0, we can gather the values of a, b, and c

a=1\\b=4\\c=8

Now, we can put these numbers into our discriminant formula

b^2-4ac\\\\(4)^2-4(1)(8)\\\\16-32\\\\-16

As the result of our discriminant is negative, this quadratic equation will have two complex solutions.

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Answer:

(3, -\frac{1}{6})

Step-by-step explanation:

We can rewrite the equation as

y = \frac{x - 3}{(x - 3)(x - 9)}

Notice that we have x - 3 in both the numerator and the denominator, so it looks like we can divide it out. However, what if x - 3 is 0? Then we would have y = \frac{0}{0 \times (x - 9)} = \frac{0}{0}, which is undefined. So although it looks like the numerator and denominator can be simplified, the resulting function we would get from simplification would not have the same behavior as this one (since such a function would be defined for x = 3, but this one is not).

A point of discontinuity refers to a particular point which is included in the simplified function, but which is not included in the original one. In this case, the point which is not included in the unsimplified function is at x = 3. In the simplified version of the function, if we plug in x = 3, we get

y = \frac{1}{((3) - 9)} = -\frac{1}{6}

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It's also important to distinguish between specific points of discontinuity and vertical asymptotes. This function also has a vertical asymptote at x = 9 (since it causes the denominator to be 0), but the difference in behavior is that in the case of the asymptote, only the denominator becomes 0 for a specific value of x

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