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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

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
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
```
Associated Topics:
High School Basic Algebra
Middle School Algebra

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