March Regional Algebra 2 Individual Solutions 1. D Isolating x on one side and solving the equation yields x = -3 2. B The minimum value of a concave up, quadratic function occurs at the vertex of the parabola. The x coordinate of a parabolas vertex is given by b2a which equals -3. Plugging this x value into the function, we see the minimum value is 11. 3. D Using the heavy side method, we get the following: A + B x +1( )x 3 = 4x 3 substituting (-1) for x, we get A = -1 and once again using the heavy side method, we get B = 1, thus the answer is -1. 4. E The maximum number of students that can be in the classroom, and have less than 15 students of each gender is 28 students: 14 males and 14 females. Thus, if there are 29 students in the classroom, there must be at least either 15 males or 15 females. 5. C The time it takes for Ari and Lisa to reach each other is (20 mph + 30 mph)(time) = 100 miles. Solving for time yields 2 hours. Since the duck flies faster than either Ari or Lisa, the distance it travels will simply equal its rate times the time it travels. Thus the duck travels (2 hours)(40 mph) = 80 miles. 6. C The balls height follows a geometric series with a common ratio of one half. After it is initially dropped from 100 feet, the ball bounces up 50 feet and then back down 50 feet, up 25 feet and then back down 25 feet, etc. So the total vertical distance it travels: 100 + 2 50

1 12

=100 + 200 = 300

7. C Contracting the right hand expression we get log215( ) log15 4( ). Using change of base we get the expression: log215( ) log2 4log215 which simplifies to 2.

8. D After substitution we get the following equation: 6 + x = x. This is now a simple quadratic equation: x2 x 6 = 0, which yields the solutions 3 and -2. Since, the solution cannot be negative, the answer is 3. 9. C We will use Cramers Rule to solve this system. Although solving via elimination will also work. The determinant for the coefficient matrix (the one you put on the denominator) is 3 2 22 1 2

1 3 3= 7 (in the system, the

last equation has the variables out of order, so be careful). The determinant for the numerator of the x value is: 4 2 21 1 2

1 3 3=14. Thus, the value of x

is 147 = 2. Substituting this value into any two of the three equations; were left with a simple two variable system of equations which can be solved via elimination. The value of y = 5, the value of z = 4. x + y + z =11 10. E Complex roots can either be imaginary or real. Since, this polynomial is of the fifth degree it therefore, has five complex roots. 11. B There are three scenarios that will satisfy the equation: a. The base quadratic equation equals 1. i. Since the determinant is greater than zero, both the roots of this quadratic are real; the sum of the roots = ba which yields 4 b. The exponent equals 0 i. Setting x2 + x 2 = 0 we solve and get the values of -2 and 1. c. The base quadratic equation equals -1, and the exponent is an even number i. Setting the base quadratic equal to -1 and solving we get: x2 4x + 2 = 1 x2 4x + 3 = 0. Solving this equation yields roots of 3 and 1. Thus, the two roots from Case A yield a sum of 4. The sum of the roots from Case B yield a sum of -1. For Case C, since 1 is already counted in the sum of the roots from Case B, the sum of the roots is just 3. 4 1+ 3 = 6

12. A The three roots form an arithmetic sequence, so let the following three expressions be the three roots: a + r, a, a r. By Vietas Formulas, the sum of the roots = ba =15. 3a =15 a = 5, once again by Vietas Formulas, the product of the roots = da (for an odd degree polynomial) = 80, 5+ r( ) 5 r( ) 5 = 80 r = 3. Thus, the roots are 2, 5, and 8. We can either FOIL out the polynomial, or use Vietas Formula for the sum of the roots taken two at a time to obtain the value of C = 66. 13. A In order to find f x( ), we need to find some function u, such that f 1+ u2 = f x( ), solving for u , we get u = 2x 2. Thus, f 1+ 2x 22 = f x( ) = 2x 2( )2 + 2x 2 + 4. So, f x( ) = 4x2 6x + 6 14. C Completing the square and putting the ellipse in standard form we get: 4 x + 2( )2191 + 3 y 5( )2191 =1. The ugly denominator is irrelevant because were searching for the center: (-2, 5) 15. B Let z = a + bi, the real portion must equal the real portion and the imaginary portion must equal the imaginary portion. Setting the imaginary portions equal to each other, we get bi = 4i, b = 4. Setting the real portions equal, we get a + a2 + 42 = 8. After subtracting a from both sides, squaring both sides to remove the square root, and solving for a , we get a = 3. Thus z = 3+ 4i and the magnitude of z = 5 16. E f 1( ) = 3, g x( ) = 29

17. C Call the desired sum S = 15 + 225 + 3125 + 4625 + 53125 + ... Therefore, 5S =1+ 25 + 325 + 4125 + 5625 + ... Subtracting these S from 5S, we get: 4S =1+ 15 + 125 + 1125 + 1625 + ... Therefore, 4S = 11 15

= 54 S =516 18. B f x( ) and h x( ) fail the horizontal line test; g x( ) is a one to one logarithmic function and therefore has an inverse 19. A An effective way to approach these problems is to draw a Venn Diagram and fill in the information in each respective area of the diagram; in this problem, there are two unknowns: the number of people that like all three fruits (lets call it x) and the number of people that like mango and grapefruit, but not strawberries (lets call it y). Setting up the following equations based on the information yields our answer: x + 504 + 73+14 = total number of Strawberry fans = 670 x = 79 x + y + 73+ 436 = total number of Grapefruit fans = 594 y = 6 20. B When a function f x( ) is divided by a linear polynomial x a( ) , the remainder of the division is equal to f a( ) - this is the logic of synthetic division. Thus, our quadratic polynomial contains the following points: f 1( ) = 4 and f 2( ) = 2. Furthermore, since the leading coefficient is one, we can setup a simple system of equations: 1+ B +C = 4 and 4 + 2B +C = 2. Solving for B and C, we see that our polynomial is: x2 + 3x 8. As aforementioned, the remainder when our polynomial is divided by x 3( ) equals f 3( ) =10 21. C C = kt2 90 = 900k k = 110C 50( ) = 250010 = 250 22. B 3+ 2( )3 =125

23. D After matrix multiplication, we find that AXB = 12 120 4 - the determinant of this matrix is 48. 24. C A common way to change base is with the division/remainder chain: 607 divided by 8 is 75 with a remainder of 7 the ones digit of the base eight number is 7. 75 divided by 8 is 9 with a remainder of 3 the eights digit of the base eight number is 3. 9 divided by 8 is 1 with a remainder of 1 the sixty-fours digit of the base eight number is 1. 1 divided by eight is zero with a remainder of 1 the five hundred twelves digit of the base eight number is 1. Our number in base eight is: 11378 25. A x = log2 5log210

= log2 5log2 5( ) 2( )( )= log2 5log2 5( )+1

1x =log2 5( )+1log2 5

=1+ 1log2 5 1 xx =

1log2 5

So, log2 5 = x1 x 26. D By definition, the description describes a hyperbola specifically, the constant length is equal to the length of the traverse axis = 2a (for an ellipse, the sum of the distances is equal to the major axis) 27. D There are two vertical asymptotes at x = -1 and x = 3; since the degree of the leading term in both the denominator and the numerator are equal, there is a horizontal asymptote. Thus the total number of vertical and horizontal asymptotes is 3. 28. B 11+ 120 = x + y11+ 120 =11+ 2 30 = x + y + 2 xy x = 5, y = 6. Thus, 2x y = 4 29. A Another way of stating the expression is the difference between x and 91 is at most 18 this literal expression translates to the formulaic expression x 91 18

30. C The exponent on the 2x term needs to be two greater than the x2 term, after setting up the appropriate algebraic equations, we see that the following term will be the constant: 64 x2( )2 2x 4 = 240