Real and Complex Equations

Date: 09/21/98 at 19:06:07
From: Cathy
Subject: AlgII/Trig - Homeschool Teacher help

Hi again. I am a homeschool teacher who has gotten "stumped" on a few 
of the problems for Algebra II/Trig. Would you please be so kind as to 
assist me?  I really appreciate your help.

 1. The greatest of three numbers equals the product of -7 and the 
    smallest number, and the remaining (middle number) is 18 more than 
    the smallest. Find the three numbers if the greatest is 3 less than 
    twice the sum of the other two numbers.

 2. Find three consecutive even integers such that the sum of the first 
    two is greater than 25 and the sum of the last two is less than 35.

 3. Simplify:  (3x^2y - 2xy) - (2xy^2 - xy + 3x^2y).

 4. Solve 3^(2x+2) = 81 for x.

 5. The height of a triangle is 4m less than twice the length of the 
    base. Find the height and the length of the base if the area of the 
    triangle is 24m^2.

 6. Solve 3x^4 - 7x^3 - 10x^2 + 28x - 8 = 0. Give the multiplicity "m" 
    for each root when m > 1.

 7. Simplify 7 / (6x-30) - (9x-6) / (2x^2 - 18x + 40) + 11 / (3x - 12).
               _______
 8. Simplify \3/ 54b^11  (the cube root of 54b^11).

 9. Simplify i^10.

10. Factor 4x^2 + 25.

11. Solve 2x^2 = 5 - 2x, using the quadratic formula and simplify.

12. For what values of k will the sum of the roots of:
 
    x^2 - (k^2 - k)x + 10 = 0 

   be equal to 6?
            ___       ______   
13. Solve \/x+3  = \4/ 12x+9   (sqrt (x+3) = fourth root (12x+9))

As always, I thank you for your help. It is not always easy homeschool 
teaching, and it is nice to know you all are there when I get stuck.
These are all things I know, but for some reason I'm unable to figure 
out. Thanks again.

Cathy

Date: 09/22/98 at 12:01:34
From: Doctor Peterson
Subject: Re: AlgII/Trig - Homeschool Teacher help

Hi, Cathy. Your list is a little intimidating, but I'll see what I can 
do for a fellow homeschooler.

1. Call the numbers A, B, and C, from largest to smallest. We have:

   A = -7 * C
   B = C + 18
   A = 2(B + C) - 3

Putting everything in the usual form:

   A      + 7C =  0
        B -  C = 18
   A - 2B - 2C = -3

That would help for the linear combination method, but I'll use the 
substitution method starting from the original equations, by plugging 
the first two into the third:

   (-7C) = 2((C + 18) + C) - 3
       3 = 7C + 2C + 36 + 2C
       3 = 11C + 36
     -33 = 11C
      -3 = C

Now we plug this back into the first two equations:

   A = -7C = 21
   B = C + 18 = 15

Finally, A, B, and C are in the right order, so it all works.
    
2. Three consecutive even integers would be 2X, 2X+2, and 2X+4, where 
   X is any integer. (The first one has to be 2 times something to be 
   even, and the next two even integers are 2 and 4 more than that.) We 
   want:

       2X + (2X+2) > 25
   (2X+2) + (2X+4) < 35

Simplifying:

   4X > 23  ==>  X > 23/4
   4X < 29  ==>  X < 29/4

So X is an integer between 5 3/4 and 7 1/4, which could be 6 or 7, and 
the three integers are either 12, 14, 16 or 14, 16, 18. Both sets 
work.

3. Multiply out the negative sign and eliminate parentheses:

   3x^2y - 2xy - 2xy^2 + xy - 3x^2y

Rearrange to group like terms:

   (3x^2y - 3x^2y) + (-2xy + xy) - 2xy^2
   -xy - 2xy^2

4. Use parentheses to clarify:

   3^(2x+2) = 81

Write 81 as a power of 3:

   3^(2x+2) = 3^4

Take log base 3 of both sides (that is, equate the exponents):

   2x+2 = 4

Solve:

   x+1 = 2
     x = 1


5. Let height and base be H and B.

      H = 2B - 4
   BH/2 = 24   (area)

Substitute first in second:

     B * (B - 2) = 24
   B^2 - 2B - 24 = 0

Use quadratic formula:

       2 +/- sqrt(4 + 96)   2 +/- 10
   B = ------------------ = -------- = 1 +/- 5 = 6 or -4
              2               2

Only the positive value is valid, so

   B = 6 m
   H = 8 m

6. I'm not sure what method you might have been given, since the 
   general method is very complicated. I would just look for a root by 
   trial and error. 2 happens to work. Then I would divide the 
   polynomial by x-2, and try 2 again just in case. That works, too, 
   so you're left with a quadratic which you can solve, and you know 
   that the root 2 has multiplicity 2. If you need more help on this 
   one, write back and tell me what the text suggests for quadrics.

7. You can simplify each polynomial by factoring out a constant:

      7     3(3x-2)          11
   ------ - ------------ + ------
   6(x-5)   2(x^2-9x+20)   3(x-4)

Now factor the quadratic, and then use the product of all the distinct 
factors in the denominators to form a common denominator:

      7     3(3x-2)         11
   ------ - ----------- + ------
   6(x-5)   2(x-4)(x-5)   3(x-4)

   7*(x-4) - 3(3x-2)*3 + 11*2(x-5)
   -------------------------------
             6(x-4)(x-5)

You can do the rest.

                 _____
8. Factor what's under the root, splitting the factors into groups of 
three:

   cbrt(2*3*3*3*b^3*b^3*b^3*b^2)
   cbrt(2 * 3^3 * b^3 * b^3 * b^3 * b^2)

Now since root(a*b) = root(a) * root(b), you can take the cubes out of 
the root:

   cbrt(2) * 3 * b * b * b * cbrt(b^2)
   3b^3 * cbrt(2b^2)

9. Since i^2 = -1,

   i^10 = i^(2*5)
        = (i^2)^5
        = (-1)^5
        = -1

10. It's a good thing you let me know you were getting into imaginary 
numbers. You can make this a difference of squares by using i^2 = -1:

   4x^2 + 25 = 2^2 * x^2 - i^2 * 5^2
             = (2x)^2 - (5i)^2
             = (2x + 5i) * (2x - 5i)

Without imaginary numbers, it can't be factored.

11. Write it in the usual form:

   2x^2 + 2x - 5 = 0

       -2 +- sqrt(2^2 - 4*2*(-5))
   x = --------------------------
                 2*2

       -2 +- sqrt(44)
     = --------------
              4

Now simplify, by writing sqrt(44) as sqrt(4 * 11) = 2 sqrt(11):

       -1 +- sqrt(11)
     = --------------
             2

12. The sum of the roots is -B = k^2 - k. Set this equal to 6:

   k^2 - k - 6 = 0

       1 +/- sqrt(1 + 24)   1 +/- 5
   k = ------------------ = ------- = 3 or -2
              2                2

13. Square both sides:

   x + 3 = sqrt(12x + 9)

Square again:

   x^2 + 6x + 9 = 12x + 9
       x^2 - 6x = 0
       x(x - 6) = 0

So x = 0 or 6.Since we squared the equation, we may have introduced 
extra roots, so we have to check these. They both work.

I did all these pretty quickly. If you can avoid sending too many 
problems at once, we would be able to give more attention to those we 
answer and maybe be more helpful. I don't like just answering problems 
without taking time to talk about what might be going wrong.

I'm curious what causes your problems, because most of these don't 
have any special trick. (The imaginary ones are a little tricky, 
though, and the word problems take a kind of thinking you may not be 
used to.) Maybe you can let me know if the answers reveal something 
specific I can help with on some of these.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   

Date: 09/22/98 at 12:35:28
From: Doctor Anthony
Subject: Re: AlgII/Trig - Homeschool Teacher help

1. I'll try this problem with 7.

   Let numbers be  a, b, c  in ascending order of magnitude.

   So:   

      7a = c
       b = 18+a
       c = 2(a+b) - 3   

   Substitute c = 7a into the last equation:
   
      7a = 2a + 2b -3
       3 = 2b - 5a

   So we have:

       b - a = 18       
      2b -5a =  3      

   Thus:

      2b - 2a = 36
      2b - 5a =  3
      -------------
           3a = 33

   So a = 11, then  b = 29 and c = 77. 
     
2. If the even integers are 2a, 2b, 2c, then:

      2a+2b > 25 and 2b+2c < 35
       a+b  > 12.5    b+c  < 17.5

   Also b = a+1, c = a+2, and so we have:

      2a+1 > 12.5        a+1+a+2 < 17.5
        2a > 11.5             2a < 14.5  
         a >  5.75             a <  7.25

   Thus a = 6 is a possibility. In this case, three consecutive even 
   integers are:

      12,  14,  16

   Also, a = 7 is a possibility, giving the three integers:

      14,  16,  18

3. Note:

   3x^2y - 2xy - 2xy^2 + xy - 3x^2y
   -xy - 2xy^2
   -xy(1 + 2y) 

4. Since 3^(2x+2) = 3^4,
             2x+2 = 4    
              x+1 = 2    

         and so x = 1.

5. Let the base be b, then h = 2b-4.

   So:

            area = (1/2)bh  = (1/2)b(2b-4) = 24

          b(b-2) = 24

   b^2 - 2b - 24 = 0

      (b-6)(b+4) = 0    

   Since b must be positive, b = 6, which implies h = 8.
   
6. We know one factor will be (3x +/- a). Try 3x +/- 1, and then try  
   x = +/-1, x = +/-2 and so on.

   This factorizes to (x+2) (x-2)^2 (3x-1) = 0.
   and the solutions are x = -2, 1/3, 2, 2.

7. Some rather tedious algebra gives:

      x - 60
   ------------
   3(x-4)(x-5)

8. Note: 54 = 2 x 3^3 and so 54^(1/3) = 3 x 2^(1/3)
   Thus, the answer is: 3 x 2^(1/3) x b^(11/3)

9. Since i^8 = +1 and i^2 = -1, i^10 = -1.


10. 4x^2 + 25 does not factorize in real numbers.
    In complex numbers it is (2x - 5i)(2x + 5i).

11. By the quadratic equation, 2x^2 + 2x - 5 = 0 has the roots:

        -2 +- sqrt(4 + 40)     -2 +- 6.6332
    x = ------------------  =  ------------
               4                    4

      = -2.1583  and  1.1583
 
12. Since the sum of the roots = - the coefficient of x, we require:

        k^2 - k = 6
         k(k-1) = 6      
    k^2 - k - 6 = 0
     (k-3)(k+2) = 0

    So values of k are -2 and 3.

13. Starting with sqrt(x+3) = (12x+9)^(1/4), raise both sides to the 
    4th power:

   (x+3)^2 = 12x + 9
    x^2 + 6x + 9 = 12x + 9 
    x^2 - 6x  =  0
    x(x-6) = 0    

    So x = 0 or x = 6.

    After squaring or raising both sides to some power, not all roots 
    will necessarily be correct. So we must check:

      x = 0 is one solution

    Check if x=6 also works:

         sqrt(x+3) = sqrt(9) = +/- 3
     (12x+9)^(1/4) = 81^(1/4) = +/-3    

    This also works.

    Thus, the solutions are x = 0, x = 6.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   

Date: 09/23/98 at 09:45:23
From: Anonymous
Subject: Re: AlgII/Trig - Homeschool Teacher help

Doctors Anthony and Peterson,

I just wanted to thank you again for the help you gave to this 
homeschool teacher. 

Cathy