## Monday, March 23, 2020

### Algebra 2 Problems of the Day (Jan 2020)

Daily Algebra 2 questions and answers.

More Algebra 2 problems.

January 2020, Part I

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

16. As θ increases from -π/2 to 0 radians, the value of cos θ will

(1) decrease from 1 to 0
(2) decrease from 0 to -1
(3) increase from -1 to 0
(4) increase from 0 to 1

Answer: (4) increase from 0 to 1
Cos(-π/2) = 0, Cos(0) = 1

17. Consider the following patterns:

I. 16, -12, 9, -6.75, …
II. 1, 4, 9, 16, …
III. 6, 18, 30, 42, …
IV. 1/2, 2/3, 3/4, 4/5, …

Which pattern is geometric?

(1) I
(2) II
(3) III
(4) IV

The terms in pattern I have a common ratio: -12/16 = -3/4; 9/-12 = -3/4
Pattern II is quadratic. The terms are squares.
Pattern III is arithmetic. Each term is 12 more than the prior term.
Pattern IV is none of these. It has neither a common ratio or common difference.

18. Consider the system below.

x + y + z = 9
x - y - z = -1
x - y + z 21

Which value is not in the solution, (x,y,z), of the system?

(1) -8
(2) -6
(3) 11
(4) 4

Combine the first two equations:

x + y + z = 9
x - y - z = -1
2x = 8
x = 4

Combine the second and third equations:

x - y - z = -1
x - y + z 21
-2z = -22
z = 11

Solve for y: (4) + y + (11) = 9
y + 15 = 9
y = -6

19. Which statement regarding polynomials and their zeros is true?

(1) f(x) = (x2 - 1)(x + a) has zeros of 1 and -a, only
(2) f(x) = x3 - ax2 + 16x - 16a has zeroes of 4 and a, only.
(3) f(x) = (x2 + 25)(x + a) has zeros of +5 and -a.
(4) f(x) = x3 - ax2 - 9x + 9a has zeros at +3 and a.

Answer: (4) f(x) = x3 - ax2 - 9x + 9a has zeros at +3 and a.
Choice (1) has factors (x + 1)(x - 1) and a zero at -1.
In choice (2), f(4) = 64 - 16a + 64 - 16a = 128 - 32a, not zero.
In choice (3), x2 + 25 cannot be factored in (x + 5)(x - 5), so they are not zeros.
Choice (4) can be factored as follows:
f(x) = x3 - ax2 - 9x + 9a
f(x) = x2(x - a) - 9(x - a)
f(x) = (x2 - 9)(x - a)
f(x) = (x + 3)(x - 3)(x - a), x = +3 or a.

20. If a solution of 2(2x - 1) = 5x2 is expressed in simplest a + bi form, the value of b is

2(2x - 1) = 5x2
4x - 2 = 5x2
5x2 - 4x + 2 = 0

x = ( -(-4) + SQRT( (-4)2 - 4(5)(2) ) / (2(5))

x = ( 4 + SQRT( 16 - 40 ) ) / (10)

x = ( 4 + SQRT( -24 ) ) / (10)

x = ( 4 + 2i SQRT( 6 ) / (10)

x = 2/5 + i SQRT( 6 ) / 5

The answer is choice (2) because they only wanted the value of b, not the entire term. Choice one includes i.