some algebra 2 problems! help please!?

Question

use whatever necessary (i.e. factoring, complete the square, quadraitc formal, etc..) to find the solutions.

A. x^6 - 64 = 0
B. 9x^4 - 24x^3 + 16x^2 = 0
C. x^4 + 2x^3 - 8x - 16 = 0
D. 36t^4 + 29t^2 - 7 = 0
E. (x^2 - x - 22)^4/3 --> as in x^2 - x - 22 is to the 4/3 power.
F. square root of x + square root of x-20 = 10
G. x^2 + 8x - 19 = 0
H. 4x^3 + 2x^2 - 24x - 12 = 0


when doing this, can you show me the work & explain how to do it?


THANK YOU SOOO MUCH!

Answer 1

Hi,


A. x^6 - 64 = 0
This is the difference of perfect squares. (It could also have been the difference of perfect cubes, but if you have a choice always do difference of perfect squares first.) x^6 is really (x³) squared. 64 is 8². So both factors start with x³ and both end with 8. One binomial has a "-" and the other has a "+".
This factors into:

(x³ - 8)(x³ + 8) = 0
Now these 2 factors are the difference and sum of perfect cubes. The general rule for these is:

1) Factor out a GCF if there is one.
2) Make 2 parentheses for a binomial and a trinomial.
3) The sign from the problem goes in the binomial. The opposite sign from what you just used goes in the trinomial, and is always followed by a "+" as the second sign in the trinomial. This is the only step where what you do is different on the sum of cubes from on the difference of cubes.
4) Figure out what you would cube to get the first term of the problem and put that expression as the first term of your binomial. Then figure out what you would cube to get the second term of the problem and put that expression as the second term of your binomial.
5) Your trinomial already has its signs, so don't worry about the signs. To find terms of the trinomial, write the first term of your binomial 3 times followed by the second term of your binomial 3 times out to the side. Multiply the first 2 things together to get the trinomial's first term. Multiply the next 2 things together to get the trinomial's middle term. Multiply the last 2 things together to get the trinomial's third term.

For x³ - 8 , there is no GCF.
Parentheses would be ( - )( + + )
To get x³, you'd cube x.
To get 8, you'd cube 2.
So your binomial becomes (x - 2)( + + )
Now write x x x 2 2 2
x*x = x²
x*2 = 2x
2*2 = 4
So with your trinomial, the factors become (x - 2)(x² + 2x + 4)
These are factors for x³ - 8.

For x³ + 8 , there is no GCF.
Parentheses would be ( + )( - + )
To get x³, you'd cube x.
To get 8, you'd cube 2.
So your binomial becomes (x + 2)( - + )
Now write x x x 2 2 2
x*x = x²
x*2 = 2x
2*2 = 4
So with your trinomial, factors are (x + 2)(x² - 2x + 4)
These are factors for x³ + 8.

So all the factors are:
(x - 2)(x² + 2x + 4)(x + 2)(x² - 2x + 4) = 0

The binomials are easy to solve.
x - 2 = 0
x = 2 <== 1st answer

x + 2 = 0
x = -2 <== 2nd answer

The other 2 factors must be solved by the quadratic formula, x = [ -b ± √(b² - 4ac)]/(2a).

For x² + 2x + 4 = 0, x = [ -2 ± √(2² - 4(1)(4))]/(2(1)) =

[ -2 ± √(-12)]/(2) =

-2 ± 2i√3
------------ =
......2
............_
-1 ± i√3 <== 3rd and 4th answers

For x² - 2x + 4 = 0, x = [ -(-2) ± √((-2)² - 4(1)(4))]/(2(1)) =

[ 2 ± √(-12)]/(2) =

2 ± 2i√3
------------ =
......2
..........._
1 ± i√3 <== 5th and 6th answers

So all 6 answers are:
2, -2,.._
-1 ± i√3,...._
AND 1 ± i√3


B. 9x^4 - 24x^3 + 16x^2 = 0

Factor out the GCF of x²:

x²(9x² - 24x + 16) = 0

Factor 9x² - 24x + 16 into (3x - 4)(3x - 4). The explanation of how to get this follows. If you don't need it, skip to *******.

To factor 9x² - 24x + 16 Look for a GCF None this time. If there was one, factor it out. Then temporarily start both parentheses with the first number and variable.
(9x.......)(9x..........) First sign goes in first parentheses.
(9x..-....)(9x..........) Product of signs goes in 2nd parentheses.
(9x..-....)(9x..-......) <== neg is because neg x pos = negative

Now multiply your first and third numbers together. Ignore their signs - you've already done them. 9 x 16 = 144 So, out to the side list pairs of factors of 144.

144
------
1, 144
2, 72
3, 48
4, 36
6, 24
8, 18
9, 16
12, 12

Now you want to pick which factors go in your parentheses, using these rules:

If the signs you put on your parentheses are the SAME, find the factors that ADD up to the middle number in the problem.
If the signs you put on your parentheses are the DIFFERENT, find the factors that SUBTRACT to the middle number in the problem. (Note you look at signs in the parentheses, not in the problem.)

(9x.-...)(9x..-....) Your signs are the same, so you want to add factors to get 24. Those factors are 12 and 12. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(9x.-.12)(9x.-.12)

Now you have to reduce either or both parentheses by dividing each parentheses' terms by the largest possible divisor. In our problem, both parentheses are divisible by 3.

(9x.-.12)(9x.-.12)
-----------..----------
.....3.............3
This reduces to your final factors of

(3x - 4)(3x - 4)

********
So factored form is x²(3x - 4)(3x - 4) = 0.
These factors are the same and give the same answer. Setting these factors equal to 0 and solving gives:

x² = 0
√x² = √0
x = 0 <== answer from x², which is a double root

3x - 4 = 0
3x = 4
x = 4/3 <== 3rd and 4th answers, which is another double root.

The solutions are 0 and 4/3.

C. x^4 + 2x^3 - 8x - 16 = 0

Now we will look at is 4 term problems. As always, look for a GCF first for the entire problem. there is none this time. Then try to factor a GCF out of just the first 2 terms and then try to factor a GCF out of the last 2 terms. If you can do this, the expression left behind both times must be the exact same expression. The 2 GCFs go together to make one factor and the repeated expression is the second factor. If either expression has an exponent in it, check to see if it can be factored again.

Given the problem: x^4 + 2x^3 - 8x - 16 = 0

There is no GCF for the entire problem.
The first 2 terms has a GCF of x³. Factor it out.

x³(x + 2) - 8x - 16 = 0

Now factor out the GCF of the last 2 terms. It will always use the sign in front of the 3rd term of the problem. So while you could divide out 8, divide out a -8 instead.

x³(x + 2) - 8(x + 2) = 0

Notice the expressions in the parentheses are exactly the same. This must always happen.

Now put x³ and - 8 together as one factor, (x³ - 8). (x + 2) is the other factor.

(x³ - 8)(x + 2) are the factors. But the first term is the difference of perfect cubes and factors again. We did this same factor in part A, so look there if you need to see the work. This factors again into:

(x - 2)(x² + 2x + 4)(x + 2) = 0

Setting the binomials equal to zero and solving gives x = 2 and -2. Solving the trinomial by the quadratic formula as we did in part A gives the answers:
..........._
-1 ± i√3

So the solutions are:
..........._
-1 ± i√3 , 2, and -2



D. 36t^4 + 29t^2 - 7 = 0

This is a trinomial. Look for a GCF. There is none this time. If there was one, factor it out. Then temporarily start both parentheses with the first number and variable to half the exponent.
(36t².......)(36t²......) First sign goes in first parentheses.

(36t².+....)(36t²......) Product of signs goes in 2nd parentheses.
(36t².+....)(36t².-...) <== neg is because pos x neg = negative

Now multiply your first and third numbers together. Ignore their signs - you've already done them. 36 x 7 = 252. So, out to the side list pairs of factors of 252.

252
------
1, 252
2, 126
3, 84
4, 63
6, 42
7, 36
9, 28
12, 21
14, 18

Now you want to pick which factors go in your parentheses, using these rules:

If the signs you put on your parentheses are the SAME, find the factors that ADD up to the middle number in the problem.
If the signs you put on your parentheses are the DIFFERENT, find the factors that SUBTRACT to the middle number in the problem. (Note you look at signs in the parentheses, not in the problem.)

(36t².+....)(36t².-...) Your signs are DIFFERENT, so you want to subtract factors to get 29. Those factors are 7 and 36. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(36t².+.36)(36t².-.7)

Now you have to reduce either or both parentheses by dividing each parentheses' terms by the largest possible divisor. In our problem, the first parentheses is divisible by 36 but the second parentheses does not reduce.

(36t².+.36)(36t².-.7)
-------------
......36
This reduces to factors of

(t².+.1)(36t².-.7) = 0

Neither of these factor again. Setting each equal to zero and solving gives:

t² + 1 = 0
t² = -1
√t² = √-1
t = ±i

36t² - 7 = 0
36t² = 7
t² = 7/36
√t² = ±√(7/36)
t = ±√(7)/6

Answers are:
......_
....√7
±.------ and ± i
......6


E. (x^2 - x - 22)^4/3 --> as in x^2 - x - 22 is to the 4/3 power.

If this is supposed to be (x^2 - x - 22)^4/3 = 0, then raise both sides to the 3/4 power. It would become:

(x^2 - x - 22)^4/3 = 0

[(x^2 - x - 22)^4/3]^(3/4) = 0^(3/4) This simplifies to:

x² - x - 22 = 0

Solving this by the quadratic formula gives:

x = [ -b ± √(b² - 4ac)]/(2a)
x = [ -(-1) ± √((-1)² - 4(1)(-22))]/(2(1)) =
x = [ 1 ± √89]/(2)

The solutions are:
...............__
x = 1 ± √89
.....------------
...........2

F. square root of x + square root of x-20 = 10
.._......____
√x + √x-20 = 10

Start by getting one term with a radical alone on one side of the equation. Here we will subtract √(x-20) on both sides.

.._..............___
√x = 10 - √x-20

Now square both SIDES of the equation. Do NOT square TERM by TERM - square the entire side.

...._.................___
(√x)² = (10 - √x-20)²

x = (10 - √(x-20))(10 - √(x-20))
x = 100 -10√(x-20) - 10√(x-20) + x - 20
x = 80 -20√(x-20) + x

Subtract x on both sides.

0 = 80 -20√(x-20)

Divide every term by 20.

0 = 4 -√(x-20)

Move the radical to the other side.
.._____
√(x-20) = 4

Square both sides again.

x - 20 = 16

Solving x = 36.

These answers always need plugged back into the original problem to see if they work, because you can get extraneous roots that need eliminated.

If x = 36, then
.._......____
√x + √x-20 = 10 becomes:

..__......____
√36 + √36-20 = 10

..__......__
√36 + √16 = 10

6 + 4 = 10

This checks, so x = 36 is the answer to this problem.




G. x^2 + 8x - 19 = 0
Solving this by the quadratic formula gives:

x = [ -b ± √(b² - 4ac)]/(2a)
x = [ -8 ± √(8² - 4(1)(-19))]/(2(1))
x = [ -8 ± √(64 + 76)]/(2)
x = [ -8 ± √(140)]/(2)
x = [ -8 ± 2√(35)]/(2)
x = -4 ± √(35) <== answers

H. 4x^3 + 2x^2 - 24x - 12 = 0
This problem has a GCF of 2 for the entire problem. Dividing it out leaves:

2[2x^3 + x^2 - 12x - 6] = 0

Now the first 2 terms inside the bracket have a GCF of x². Divide it out of those terms, but keep it inside the bracket.

2[x²(2x + 1) - 12x - 6] = 0

The next 2 terms have a GCF of -6 because this GCF always uses the sign of the thirds term. This factors into:

2[x²(2x + 1) - 6(2x + 1)] = 0

The 2 GCFs inside the bracket combine to become one factor while the repeated factor is the other. With the GCF of 2 out front, factors are:

2(x² - 6)(2x + 1) = 0

Factors with variables can be solved for x, so we can ignore the GCF of 2.

x² - 6 = 0
x² = 6
x = ±√6

2x + 1 = 0
2x = -1
x = -½


I hope those help!! I tried to include your explanations!! :-)

Answer 2

a) This is the difference of 2 squares so it factors to sqrt of 1st term - sqrt 2nd times sqrt 1st + sqrt 2nd like this:

(x^3 - 8)(x^3+8)=0

Now this factors further because we have a sum and difference of cubes:
(x - 2)(x² + 2x + 4)(x + 2)(x² - 2x + 4) = 0

so to equal zero any one or combination of factors or all have to equal zero so:
x=2 OR x=-2 OR use the quadratic equation on the second and fourth terms to get:
x = -1 ± i√3


b) Here you factor out x^2 to get:
x^2 (9x^2 - 24x +16) = 0
and we have two complete squares and the second can be factored to get:
x^2 (3x-4)(3x-4) = x^2 (3x-4)^2 =0
so
x =0 and x=0 OR 3x-4=0 x=4/3 twice like te zeros

c) This one you break up into pairs and factor to get:
x^3(x+2) - 8(x+2) = 0 Now x+2 is common to both so:
(x^3 - 8)(x+2) = 0
this is like a above with a difference of cubes

(x - 2)(x² + 2x + 4)(x+2)=0

x+2=0 x=-2 OR x - 2 = 0 so x=2 OR use qudratic to obtain the same result as part a above for the middle expression.

d) Here you just factor
(36t^2-7)(t^2+1)=0
36t^2 = 7 t=6/√7 or t^2 = -1 t=√-1 = i

e) this has no solution as it does not equal anythig. so it can only be factored or simplified and it is in its simplest form.

f) √x + √(x-20) = 10
Squaring both sides gives:
x + 2 √(x^2 - 20x) + x -20 = 100
2x + 2√(x^2-20x) = 120
arrange terms to get
√(x^2-20x) = (120-2x)/2 = 60 - x
squre again to get:
x^2 - 20 x = 3600 - 120x + x^2
100x = 3600
x=36

g) use the quadratic for this one since 19 does not factor.
I think someone explained it well further down.
a=1
b=8
c=-19
and solve.

h) 2x^2 (2x+1) - 12(2x+1) = 0
(2x^2 - 12) (2x+1) = 0
2x^2 = 12 x^2 = 6 x=±√6 OR
2x+1 = 0
2x = -1
x = -1/2

Answer 3

1 x^6=64 so x=6th root of 64 x=2 (find sqrt and then cube rt)

2 take x^2 common and solve the quadratic
we get (3x-4)^3 so roots are 4/3,4/3,0,0

3 we can factorize (x+2)(x^3-8)=0

4 put t^2=y solve the quad and get the value of y and then t

5 take powers of 3/4 on both sides .. since rhs is 0 it is a simple quad

6 take x-20=y^2 .. u get sqrt (y^2+20) =10-y
square both sides and solve.

7 in completing squares,to a^2+2ab+b^2 center term is 2ab where a=1 so b has to be 4
so we have to get x^2 +8x+16=35
now solve this