More Algebra 2 problems.

__August 2017, Part I__

All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.

*1. The function f(x) = (x − 3) / (x ^{2} + 2x - 8) is undefined when x equals
1) 2 or − 4
2) 4 or − 2
3) 3, only
4) 2, only
*

**Answer: 1) 2 or − 4**

A fraction is undefined when the denominator is equal to zero, so you need to know for what values of x is

(x^{2} + 2x - 8) = 0?

Factor the polynomial into (x + 4)(x - 2) = 0

The zeroes of this equation are -4 and 2, so f(x) is undefined at -4 and 2.

*2. Which expression is equivalent to (3k − 2i) ^{2}, where i is the imaginary unit?
1) 9k^{2} − 4
2) 9k^{2} + 4
3) 9k^{2} − 12ki − 4
4) 9k^{2} − 12ki + 4
*

**Answer: 3) 9k ^{2} − 12ki − 4**

(3k − 2i)

^{2}= (3k - 2i)(3k - 2i) = 9k

^{2}- 6ki - 6ki + 4i

^{2}

= 9k

^{2}- 12ki - 4

*3. The roots of the equation x ^{2} + 2x + 5 = 0 are
1) -3 and 1
2) -1, only
3) -1 + 2i and -1 - 2i
4) -1 + 4i and -1 - 4i
*

**Answer: 3) -1 + 2i and -1 - 2i**

Using the Quadratic Formula, a = 1, b = 2, and c = 5.

The discriminant, b^{2} - 4ac, would be (2)^{2} - 4(1)(5) = 4 - 20 = -16.

This eliminates choices 1 and 2, as a negative discriminant means that there are no real roots to the equation.

x = (-b __+__ (-16)^.5) / (2(a)) = (-2 __+__ 4i ) / (2(1)) = -1 __+__ 2i

Comments and questions welcome.

More Algebra 2 problems.

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