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1 Math Xa Algebra Practice Problems (Solutions) Fall 2008 Directions Please read carefully! You will not be allowed to use a calculator or any other aids on the Algebra PreTest or PostTest. Be sure to write neatly illegible answers will receive little or no credit. You will have one hour to complete the each test, which will consist of questions. Good luck! You can review for material tested below by looking at the corresponding review materials online. Solving linear equations 1. Solve P = a + art for t. t = P a ar 2. Solve A = 2lw + 2lh + 2w for h. A 2lw 2w = 2lh A 2lw 2w = h 2l A = 2lw + 2lh + 2w Solving quadratic equations 3. Solve = 0. Use quadratic formula. The solution is = 1 ± 15/3
2 4. Solve 2/3 1/3 6. This is a quadratic equation in disguise. Let u = 1/3. Then the equations becomes u 2 u 6 = 0 Factor to find the solutions u = 2 and u = 3. Then solve 1/3 = 2 by cubing both sides to get = 8 and solve 1/3 = 3 by cubing to get = 27. Solving polynomial equations by factoring 5. Solve p 3 = pmn 2 for p. p 3 = pmn 2 p 3 pmn 2 = 0 p(p 2 mn 2 ) = 0 From here we get the solution p = 0 and the equation p 2 mn 2 = 0 which we can solve either using the quadratic formula or rewriting the equation as p 2 = mn 2 so that p = ± mn. Thus there are three solutions p = 0 and p = ± mn. Solving equations involving rational epressions 6. Solve + 2(+3) 2 5 = 2 Page 2
3 (+3) (+3) 2 5 = 2 = ( + 3) = = 2 10 = 2(9 + 12) 8 = 24 = 3 7. Solve = The common denominator here is (4 2 + )( 1) = (4 + 1)( 1) = 0 ( 1) (4 2 + )( 1) + 2(4 2 + ) ( 1)(4 2 + ) 3( 1) (4 + 1)( 1) = 0 ( 1) + 2(4 2 + ) 3( 1) = 0 (4 + 1)( 1) ) = 0 (4 + 1)( 1) (4 + 1)( 1) = 0 2(3 + 2) (4 + 1)( 1) = 0 Remember that a rational epression p()/q() is equal to zero for that make p() = 0 and q() 0. The numerator of our fraction is equal to zero at = 0 and = 2/3. But = 0 also makes the denominator zero, so this cannot be a solution. Thus the only solutions is = 2/3. Page 3
4 Solving linear inequalities 8. Solve the inequality > 2 3. > Solve the inequality 3 10 < < 9 2 or ( 3 4, 9 2 ] Solving polynomial inequalities 10. Solve the inequality > 0. First we factor the lefthand side to get ( 4)( + 5) > 0. So the lefthand side is equal to zero when = 4 and = 5. We know that ( 4)( + 5) is positive if and only if either both factors are positive or both factors are negative. Both factors are positive if > 4 and > 5 these two inequalities are both true if > 4. Both factors are negative if < 4 and < 5 these two inequalities are both true if < 5. Thus we have two regions on which our inequality is satisfied: > 4 or < 5, which we can also write as (, 5) (4, ). We can do this same analysis by noting that = 4 and = 5 are the only places where = 0 and since f() = is a continuous function we know that for all values of in ( infty, 5), f() has the same sign. To find the sign for this region, we just test a point f( 6) = 10 > 0, hence the epression is positive on (, 5). For the region ( 5, 4) we can test = 0 giving f(0) = 20 < 0 so that the epression is negative on this region. And for the region (4, ) we test f(5) = 10 > 0 and so the epression is positive on this region. This gives the same solution as above. 11. Solve the inequality 4 < 2. Page 4
5 First we get everything over to one side (and zero on the other side) and then we factor. 4 2 < 0 2 ( 2 1) < 0 2 ( 1)( + 1) < 0 4 < 2 Thus 4 2 = 0 when = 0, 1, or 1. This splits the number line into four regions on which the sign of the epression 4 2 is constant: (, 1), ( 1, 0), (0, 1), (1, ). We test points on each of these regions to find that 4 2 < 0 on the regions ( 1, 0) and (0, 1). Note that since we want the epression to be strictly less than zero, we cannot combine these two regions together! The solution is ( 1, 0) (0, 1). Solving rational inequalities 12. Find the solution to the inequality > 0 The epression is zero when = 2 and undefined when = 1. This splits the number line into three regions on which the sign of 2 is constant: (, 1), ( 1, 2), (2, ). Testing +1 points in each of these regions, we find that 2 is positive in the regions (, 1) +1 and (2, ). The solution is (, 1) (2, ). Note that 2 is not a solution to the inequality because it makes the epression zero, whereas 1 is not a solution to the inequality because it makes the epression undefined. 13. Find the solution to the inequality First, we get everything over to the lefthand side, with zero on the right Page 5
6 hand side. Net, we get a common denominator, in this case ( 4) The numerator is zero when = 7 and the denominator is zero when = 4. This splits the number line into three regions on which the sign of 3+21 is constant: 4 (, 4), (4, 7), (7, ). Testing points in each of these regions we find that 3+21 is 4 negative on the intervals (, 4) and (7, ). Since = 7 makes the epression zero and = 4 makes the epression undefined, we have the solution (, 4) [7, ). Solving absolutevalue inequalities 14. Solve the inequality t Remember that for any mathematical epression A(t), A(t) > c means that either A(t) < c or A(t) > c (and similarly A(t) < c means that c < A(t) < c). Thus in this case we have two inequalities t or t The first inequality has solution t 3 5, the second has solution t. The full 2 2 solution set is t or t, that is t (, ] [ 3, ) Solve the inequality + r c, where c is a positive real number. Page 6
7 Remember that for any mathematical epression A(t), A(t) < c means that c < A(t) < c (and similarly A(t) > c means that either A(t) < c or A(t) > c). Thus in this case we have the combination inequality c + r c c r c r Simplifying epressions involving eponents and radicals 16. Simplify the following epression as completely as possible: ( c3 d 2 ) 5 3d ( c3 d 2 ) 5 3d = d 3 c 3/2 It would also be acceptable to give the answer ( c3 d 2 ) 5 3d = /2 d 3 c 3/2 17. Simplify the following epression as completely as possible: c 3 d 2 4cd 5 d 2 Any of the following three forms would be acceptable solutions c 3 d 2 4cd 5 d 2 = c 3 d 4 4cd 3 = c 3 d 4 4c d 3 = c3 d 7 4c d 3 Page 7
8 Simplifying polynomial epressions 18. Factor the following epression as completely as possible z 2 y 9 + 2z 3 y 8 + z 4 y 7 z 2 y 9 + 2z 3 y 8 + z 4 y 7 = z 2 y 7 (y 2 + 2zy + z 2 ) = z 2 y 7 (y + z) Epand and then simplify as much as possible 2 ( ) ( 21 11) ( ) ( 21 11) = = = 2 ( ) Either of the last two lines would be considered a complete simplification. Simplifying rational epressions 20. Simplify the following + y 2 + y 2 First, we reinterpret the negative eponents as fractions. Net we try to get rid of the fractions within fractions by clearing denominators. Then we Page 8
9 simplify if possible (in this case there is no further simplification possible). + y 2 + y 2 = + y y 2 = + y y 2 2 y2 2 y 2 = ( + y)(2 y 2 ) y Simplify the following assuming 0. ( ) ( ) ( 4 7 ) ( ) + 3 = 2 3 2( + 3) Simplify the following, assuming We can use a strategy of combining the numerator into one fraction and combining the denominator into one fraction, and then instead of division we invert and multiply = = = 1 2 Page 9
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