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Algebra Practice Problems for Precalculus and Calculus

Solve the following equations for the unknown x:

1. 5=7x-16

2. 2x-3=5-x

3. 12(x-3)+x=17+3(4-x)

4. 5x=2x-3

Multiply the indicated polynomials and simplify. 5. (4x-1)(-3x+2)

6. (x-1)(x2+x+1)

7. (x+1)(x2-x+1)

8. (x-2)(x+2)

9. (x-2)(x-2)

10. (x3+2x-1)(x3-5x2+4)

Find the domain of each ofthe following functions in 11-15. 11. f(x)=1+x

12. f(x)=11+x

13. f(x)=1x

14. f(x)=11+x

15. f(x)=11+x2

16. Given that f(x)=x2-3x+4, find and simplify f(3), f(a), f(-t) , and f(x2+1) .

Factor the following quadratics 17. x2-x-20

18. x2-10x+21

19. x2+10x+16

20. x2+8x-105

21. 4x2+11x-3

22. -2x2+7x+15

23. x2-2

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Solve the following quadratic equations in three ways: 1) factor, 2) quadratic formula, 3) complete the square 24. x2+6x-16=0

25. -x2-3x-2=0

26. 2x2+2x-4=0

Solve the following smorgasbord of equations and inequalities 27. x=2x-1

28. x2-3=2x

29. |x-5|=4

30. 2x+4>3

31. -2x+4>3

32. x+4x-3=2

33. x2-x-2>0

Add/Subtract the following rational expressions:

34. xx+2+3x-4

35. x2+1(x-1)(x-2)-x3x-3

Simplify the following rational expressions (if possible): 36. x2+x-2x2-1

37. x2+5x+6x2-3x+2

38.

xx+2+3x+1x-1

Solutions

1. Given that 5=7x-16, add 16 to both sides to get 7x=21. Now divide both sides by 7 to get x=3. Checking, we see that 7 (3) -16=21-16=5.

2. Given that 2x-3=5-x, add x to both sides and then add 3 to both sides to get 3x=8. Now divide both sides by 3 to get x=8/3=2.6¯. Checking, we see that 2(8/3)-3=163-3=163-93=73 and 5-(8/3)=153-83=73.

3. Given that 21(x-3)+x=17+3(4-x) , we first simplify the left and right hand sides using the distributive property to get 2X1-23+x=17+12-3x. Combining like terms on both sides gives 2X3-23=29-3x. Now we add 3x and 23 to both sides, obtaining 2X9=612. Dividing both sides by 29 (or multiplying both sides by 92) gives x=612 92=619. Checking we see that LHS=12 (619-279)+619=12349+619=179+619=789 and RHS=17+3 (369-619)=17+3(-259)=17-759=

1539-759=789.

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4. Given that 5x=2x-3, we ""cross-multiply“ to obtain 5(x-3)=2x. Distributing the 5 gives 5x-15=2x. Subtracting 2x and adding 15 to both sides gives 3x=15. Dividing both sides by 3 gives x=5. Checking, we see that 55=1 and 25-3=22=1.

NOTE: Checking is very important in this kind ofproblem. When there are x's in the denominator of fr actions in equations, it is possible that your final ""solution“ doesn't satisfy the original equation (because you would divide by zero) so it is really not a solution to the original problem.

In the ""multiply and simplify“ problems, we must multiply each term in the left-hand factor with each term in the right-hand factor, and then simplify by combining like terms. In the case of binomial × binomial, we can use the so-called FOIL method. 5. (4x-1)(-3x+2)=-12x2+8x+3x-2=-12x2+11x-2.

6. (x-1)(x2+x+1)=x3+x2+x-x2-x-1=x3-1.

7. (x+1)(x2-x+1)=x3-x2+x+x2-x+1=x3+1.

8. (x-2)(x+2)=x2+2x-2x-4=x2-4.

9. (x-2)(x-2)=x2-2x-2x+4=x2-4x+4.

10. (x3+2x-1)(x3-5x2+4)=x6-5x5+4x3+2x4-10x3+8x-x3+5x2-4=x6-5x5+2x4-7x3+5x2+8x-4 11. For a number x to be in the domain ofthe function f(x)=1+x, we require 1+x>0 (so we don't take the square root ofa negative number). Subtracting 1 fr om both sides ofthis inequality yields x>-1. Thus, in interval notation, the domain is the set [-1, ) .

12. For a number x to be in the domain ofthe function f(x)=11+x, we require 1+x0 (so we don't divide by zero). Thus, we require x-1. In interval notation, the domain is the set (, -1)(-1, ) .

13. For a number x to be in the domain ofthe function f(x)=1x, we require x>0 (so we don't divide by zero or take the square root of a negative number). In interval notation, the domain is the set (0, ) .

14. For a number x to be in the domain ofthe function f(x)=11+x, we require 1+x>0 (so we don't divide by zero or take the square root of a negative number). Subtracting 1 fr om both sides ofthis inequality yields x>-1. Thus, in interval notation, the domain is the set (-1, ) .

15. For a number x to be in the domain ofthe function f(x)=11+x2, we require that 1+x20 (so we don't divide by zero). But no matter what x is, 1+x2>0. Therefore, the domain is R, the set of all real numbers.

16. f(3)=(3)2 -3(3) +4=9-9+4=4, f(a)=a2-3a+4, f(-t)=(-t)2-3(-t)+4=t2+3t+4, and

f(x2+1)=(x2+1)2-3(x2+1)+4=x4+2x2+1-3x2-3+4=x4-x2+2.

By trial & error, we see that:

17. x2-x-20=(x-5)(x+4)

18. x2-10x+21=(x-3)(x-7)

19. x2+10x+16=(x+8)(x+2)

20. x2+8x-105=(x+15)(x-7)

21. 4x2+11x-3=(4x-1)(x+3)

22. -2x2+7x+15=-(2x2-7x-15)=-(2x+3)(x-5)

23. x2-2=(x-2)(x+2) tricky, tricky, tricky!!! :)

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24. x2+6x-16=0

(a) By factoring: x2+6x-16=0 implies that (x+8)(x-2)=0. Thus, x=-8 or x=2.

(b) By the quadratic formula:

x=-b±b2-4ac2a=-6±36-4(1)(-16)2(1)=-6±1002=-6±102=2 or -8

(c) By completing the square: by adding 16 to both sides of x2+6x-16=0, we get x2+6x=16. Now, adding 9=32=(6/2)2 to both sides makes the left-hand side a perfect square: x2+6x+9=25. We can factor the left hand side to get (x+3)2=25. Now take the square root of both sides, allowing for the two square roots of25 on the right hand side to obtain x+3=±5. Subtracting 3 fr om both sides gives x=-3±5. In other words, x=-8 or x=2.

Checking: (-8)2+6(-8)-16=64-48-16=0 and (2) +6(2)-16=4+12-16=0.

25. -x2-3x-2=0

(a) By factoring: -x2-3x-2=0 implies that x2+3x+2=0 (multiply both sides by -1). Factoring gives (x+1)(x+2)=0. Thus, x=-1 or x=-2.

(b) By the quadratic formula:

x=-b±b2-4ac2a=3±9-4(-1)(-2)2(-1)=3±1-2=3±1-2=-2 or -1

(c) By completing the square: by adding 2 to both sides of -x2-3x-2=0, we get -x2-3x=2. Now, multiply both sides by-1 to get x2+3x=-2. Now, adding 9/4=(3/2)2 to both sides makes the left-hand side a perfect square: x2+3x+94=-2+94=14. We can factor the left hand side to get (x+32)2=14. Now take the square root of both sides, allowing for the two square roots of1/4 on the right hand side to obtain x+32=±12. Subtracting 3/2 from both sides gives x=-32±12. In other words, x=-1 or x=-2.

Checking: -(-1)2-3(-1)-2=-1+3-2=0 and -(-2)2-3(-2)-2=-4+6-2=0. 26. 2x2+2x-4=0

(a) By factoring: 2x2+2x-4=0 implies that x2+x-2=0 (divide both sides by 2). Factoring gives (x+2)(x-1)=0. Thus, x=-2 or x=1.

(b) By the quadratic formula:

x=-b±b2-4ac2a=-2±4-4(2)(-4)2(2)=-2±364=-2±64=1 or -2

(c) By completing the square: by adding 4 to both sides of 2x2+2x-4=0, we get 2x2+2x=4. Now divide both sides by 2 to give us x2+x=2. Now, adding 14=(1/2)2 to both sides makes the left-hand side a perfect square: x2+x+14=2+14=94. We can factor the left hand side to get (x+12)2=94. Now take the square root of both sides, allowing for the two square roots of 49 on the right hand side to obtain x+21=±32. Subtracting 1/2 from both sides gives x=-12±32. In other words, x=1 or x=-2.

Checking: 2 (1) +2(1)-4=2+2-4=0 and 2 (-2)2+2(-2)-4=8-4-40=0.

27. Squaring both sides of x=2x-1 gives x=2x-1. Solving this equation for x gives us x=1. Checking in the original equation: LHS=1=1, RHS=2(1)-1=1=1.

28. Squaring both sides of x2-3=2x gives x2-3=2x. Subtract 2x fr om both sides to get x2-2x-3=0. The left-hand

side can now be factored to give (x-3)(x+1)=0, so x=3 or x=-1. CHECKING in the ORIGINAL equation:

LHS=(3)2-3=9-3=6=2(3)=RHS, however, ifyou plug x=-1 into either side ofthis equation, you get the square root ofa negative number. Therefore, for us, x=-1 is not a solution (even though the LHS and RHS are equal the imaginary number -2=j2).

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29. |x-5|=4 implies that either x-5=4 or x-5=-4. Thus, either x=9 or x=1. Checking shows that both of these numbers are solutions: |9-4|=|4|=4 and |1-5|=|-4|=4.

30. Subtracting 4 fr om both sides of2x+4>3 gives 2x>-1. Now divide both sides by 2 to get x>-12. The logic also works in the other direction: if x>-12, this will imply that 2x+4>3. Thus, the solution set is the interval [hyphen 12, ).

31. Subtracting 4 fr om both sides of -2x+4>3 gives -2x>-1. Now divide both sides by -2 and switch the direction of the inequality to get x<12. The logic also works in the other direction: if x<12, this will imply that -2x+4>3. Thus, the solution set is the interval (, 12].

32. Take the equation x+4x-3=2 and multiply both sides by x-3 to get x+4=2(x-3) . Now solve this equation: x+4= 2x-6x=10. Checking: 10+410-3=14/7=2.

33. x2-x-2>0 implies that (x-2)(x+1)>0. Thus, either x-2>0 and x+1>0 OR x-2<0 and x+1<0. Thus, either x>2 and x>-1 OR x<2 and x<-1. Thus, either x>2 OR x<-1. The logic also works in the other direction: if x>2 OR x<-1, then (x-2)(x+1)>0 so x2-x-2>0. Thus, the solution set is (, -1)(2, ) .

34. We need to get a common denominator. The simplest one to choose is (x+2)(x-4) . Multiplying the top and bottom ofthe first fr action by x-4 and multiplying the top and bottom ofthe second fraction by x+2 and combining the fr actions produces:

xx+2+3x-4=x(x-4)(x+2)(x-4)+3(x+2)(x+2)(x-4)=(x2-4x)+(3x+6)(x+2)(x-4)

Simplify ing the top and bottom gives us:

x2-x+6

x2-2x-8

Ifthere were a common factor on the top and bottom, we would cancel it out. However, there are no common factors. Therefore, this is our final answer.

35. We need to get a common denominator. The simplest one to choose is (x-1)(x-2)(x-3) . Multiplying the top and bottom ofthe first fraction by x-3 and multiplying the top and bottom of the second fr action by (x-1)(x-2) and combining the fractions produces the following expressions (which are equal to the original):

(x2+1)(x-3)(x-1)(x-2)(x-3)-x3(x-1)(x-2)(x-1)(x-2)(x-3)=(x3-3x2+x-3)-(x5-3x4+2x3)(x-1)(x-2)(x-3)

Simplify ing the top and bottom gives us:

-x5+3x4-x3-3x2+x-3(x-1)(x2-5x+6)=-x5+3x4-x3-3x2+x-3x3-6x2+11x-6

Again, we would technically need to check for common factors to simplify this ""completely“. To do this, it is enough to determine whether 1, 2, or 3 are zeros of the polynomial in the numerator (since they are the zeros of the polynomial in the denominator). Let's call the numerator P, so P(x)=-x5+3x4-x3-3x2+x-3. P(1)=-1+3-1-3+1-3=-40, P(2)=-32+48-8-12+2-3=-50, and P(3)=-243+243-27-27+3-3=-540. Therefore, there a no common factors, so the answer above is as simple as possible.

36. We can factor the top and bottom to obtain

x2+x-2x2-1=(x+2)(x-1)(x+1)(x-1)

There is a common factor ofx-1, so we can cancel this to obtain our final answer:

x+2x+1

It should be pointed out that this expression is equal to the first as long as x1.

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37. We factor the numerator and denominator to obtain:

x2+5x+6x2-3x+2=(x+3)(x+2)(x-1)(x-2)

The numerator and denominator here have no common factors. Thus, this expression cannot be simplified.

38. The quickest approach here is to immediately multiply the top and bottom ofthis ""double-decker fr action“ by (x+2)(x-1) . Doing this, and canceling things, we obtain:

(xx+2+3)x+1x-1(x+2)(x-1)(x+2)(x-1)=x(x-1)+3(x+2)(x-1)(x+1)(x+2)

Simplifying:

x2-x+3x2+3x-6x2+3x+2=4x2+2x-6x2+3x+2

The top ofthis can be factored as 2(2x+3)(x-1) , and so has no common factors with the bottom. Therefore, we are done.

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