Assignment 12

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course Math 158

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

012. `* 12

* 1.4.12 (was 1.4.6). Explain how you found the real solutions of the equation (1-2x)^(1/3) - 1 = 0

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Your solution: You will add 1 to both sides to get. (1-2x)^(1/3) = 1. You will raise both sides to the power of 3. You will have 1 -2x = 1. You will add -1 to both sides. You have -2x = 0. You divide by -2. Your answer is x = 0.

confidence rating #$&*: 3

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Given Solution:

* * Starting with

(1-2x)^(1/3)-1=0

add 1 to both sides to get

(1-2x)^(1/3)=1

then raise both sides to the power 3 to get

[(1-2x)^(1/3)]^3 = 1^3.

Since [(1-2x)^(1/3)]^3 = (1 - 2x) ^( 1/3 * 3) = (1-2x)^1 = 1 - 2x we have

1-2x=1.

Adding -1 to both sides we get

-2x=0

so that

x=0.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: * 1.4.28 (was 1.4.18). Explain how you found the real solutions of the equation sqrt(3x+7) + sqrt(x+2) = 1.

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Your solution: You will subtract sqrt(x +2) from both sides. You will have sqrt(3x +7) = -sqrt(x+2) +1. We square both sides. You will have (3x +7)^2 = (sqrt(x+2) +1)^2. You will have 3x + 7= x + 2 - 2sqrt (x+ 2) + 1. Simplify: 3x + 7 = x + 3 - 2sqrt(x +2). You will subtract x + 3 from both sides. You get 3x + 7 - x - 3 = -2sqrt (x + 2). You will collect like terms: 2x + 4 = -2sqrt (x +2). You will square both sides to get: 4x^2 + 16x + 16 =4x + 8. You will subtract 4x + 8 from both sides to get 4x^2 +12x+ 8 = 0. You will factor to get: 4(x^2 + 3x + 2)= 0. Factor totally: 4(x+2)(x+1) = 0. You will divide by 4. You will make two problems. x + 2 = 0 so x = -2. Problem 2: x + 1 = 0. x = -1. Your solution is -2, -1.

confidence rating #$&*: 3

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Given Solution:

* * Starting with

sqrt(3x+7)+sqrt(x+2)=1

we could just square both sides, recalling that (a+b)^2 = a^2 + 2 a b + b^2.

This would be valid but instead we will add -sqrt(x+2) to both sides to get a form with a square root on both sides. This choice is arbitrary; it could be done either way. We get

sqrt(3x+7)= -sqrt(x+2) + 1 .

Now we square both sides to get

sqrt(3x+7)^2 =[ -sqrt(x+2) +1]^2.

Expanding the right-hand side using (a+b)^2 = a^2 + 2 a b + b^2 with a = -sqrt(x+2) and b = 1:

3x+7= x+2 - 2sqrt(x+2) +1.

Note that whatever we do we can't avoid that term -2 sqrt(x+2).

Simplifying

3x+7= x+ 3 - 2sqrt(x+2)

then adding -(x+3) we have

3x+7-x-3 = -2sqrt(x+2).

Squaring both sides we get

(2x+4)^2 = (-2sqrt(x+2))^2.

Note that when you do this step you square away the - sign. This can result in extraneous solutions.

We get

4x^2+16x+16= 4(x+2).

Applying the distributive law we have

4x^2+16x+16=4x+8.

Adding -4x - 8 to both sides we obtain

4x^2+12x+8=0.

Factoring 4 we get

4*((x+1)(x+2)=0

and dividing both sides by 4 we have

(x+1)(x+2)=0

Applying the zero principle we end up with

(x+1)(x+2)=0

so that our potential solution set is

x= {-1, -2}.

Both of these solutions need to be checked in the original equation sqrt(3x+7)+sqrt(x+2)=1

As it turns out:

the solution -1 gives us sqrt(4) + sqrt(1) = 1 or 2 + 1 = 1, which isn't true,

while

the solution -2 gives us sqrr(1) + sqrt(0) = 1 or 1 + 0 = 1, which is true.

• x = -1 is an extraneous solution that was introduced in our squaring step.

• Thus our only solution is x = -2. **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: * 1.4.40 (was 1.4.30). Explain how you found the real solutions of the equation x^(3/4) - 9 x^(1/4) = 0.

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Your solution: We factor x^1/4 from both sides. This gives us x^(1/4)(x^(1/2) -9 = 0. We break this into 2 problems. Problem 1: x^(1/4) = 0. This is x = 0. Problem 2: x^(1/2) - 9 = 0. You will add 9 to both sides to get: x^(1/2) = 9. You will square both answers to get: 0, 81.

confidence rating #$&*: 3

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Given Solution:

* * Here we can factor x^(1/4) from both sides:

Starting with

x^(3/4) - 9 x^(1/4) = 0

we factor as indicated to get

x^(1/4) ( x^(1/2) - 9) = 0.

Applying the zero principle we get

x^(1/4) = 0 or x^(1/2) - 9 = 0

which gives us

x = 0 or x^(1/2) = 9.

Squaring both sides of x^(1/2) = 9 we get x = 81.

• So our solution set is {0, 81). **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: * 1.4.46 (was 1.4.36). Explain how you found the real solutions of the equation x^6 - 7 x^3 - 8 =0

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Your solution: I am confused as to why you substituted the a. I know how to do this on the calculator by using y = and 2nd graph to get the solution (-1, 2)

confidence rating #$&*: Lost

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Given Solution:

* * Let a = x^3.

Then a^2 = x^6 and the equation x^6 - 7x^3 - 8=0 becomes

a^2 - 7 a - 8 = 0.

This factors into

(a-8)(a+1) = 0,

with solutions

a = 8, a = -1.

Since a = x^3 the solutions are

• x^3 = 8 and

• x^3 = -1.

We solve these equations to get

• x = 8^(1/3) = 2

and

• x = (-1)^(1/3) = -1.

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Self-critique (if necessary):

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Self-critique Rating:

@&
If you substitute a for x^3, then you end up with a quadratic equation that can be easily factored.

If a = x^3, then x^6 = a^2 so the equation becomes

a^2 - 7 a - 8 = 0.

We factor this and find that a can be either 8 or -1.

So x^3 can be either 8 or -1.

Thus x can be either 2 or -1.
*@

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Question: * 1.4.64 (was 1.4.54). Explain how you found the real solutions of the equation x^2 - 3 x - sqrt(x^2 - 3x) = 2.

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Your solution: I still do not understand using u. I can do it from the 2nd step. Problem: u^2 - u - 2 = 0. Factor: (u-2)(u + 1) = 0. You get u = 2, -1. You will solve x^2 - 3x -4. Factor will be (x - 4)(x + 1) = 0. Solutions are 4, -1.

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The left-hand side consists of

x^2 - 3x

and

the square root of x^2 - 3x.

So instead of

x^2 - 3 x - sqrt(x^2 - 3x)

we write the left-hand side as

u - sqrt(u),

which is easier to deal with.

We solve for u, then come back and figure out what value(s) of x give us our values of u.
*@

confidence rating #$&*: 2

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Given Solution:

* * Let u = sqrt(x^2 - 3x).

Then u^2 = x^2 - 3x, and the equation is

u^2 - u = 2.

Rearrange to get

u^2 - u - 2 = 0.

Factor to get

(u-2)(u+1) = 0.

• Solutions are u = 2, u = -1.

Substituting x^2 - 3x back in for u we get

sqrt(x^2 - 3 x) = 2

and

sqrt(x^2 - 3 x) = -1.

The second is impossible since sqrt can't be negative.

The first gives us

sqrt(x^2 - 3x) = 2

x^2 - 3x = 4.

Rearranging we have

x^2 - 3x - 4 = 0

so that

(x-4)(x+1) = 0

and

x = 4 or x = -1.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: * 1.4.92 \ 90 (was 1.4.66). Explain how you found the real solutions of the equation x^4 + sqrt(2) x^2 - 2 = 0.

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Your solution: I am not sure how to work this one.

confidence rating #$&*: 0

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Given Solution:

* * Starting with

x^4+ sqrt(2)x^2-2=0

we let u=x^2 so that u^2 = x^4 giving us the equation

u^2 + sqrt(2)u-2=0

Using the quadratic formula

u=(-sqrt2 +- sqrt(2-(-8))/2

u=(-sqrt2+-sqrt10)/2

Note that u = (-sqrt(2) - sqrt(10) ) / 2 is negative, and u = ( -sqrt(2) + sqrt(10) ) / 2 is positive.

u = x^2, so u can only be positive. Thus the only solutions are the solutions to the equation come from

x^2 = ( -sqrt(2) + sqrt(10) ) / 2.

The solutions are

x = sqrt( ( -sqrt(2) + sqrt(10) ) / 2 )

and

x = -sqrt( ( -sqrt(2) + sqrt(10) ) / 2 ).

Approximations to three significant figures are

• x = .935

and

• x = -.935.

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Self-critique (if necessary):

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Self-critique Rating:

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&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

Add comments on any surprises or insights you experienced as a result of this assignment.

I had problems with the substitution parts of the assignments. I am used to doing this b

Self-critique (if necessary):

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Self-critique rating:

@&
Check my notes and let me know if you have additional questions.

Note that self-critiques and questions should go under self-critique, which is where I look for questions. Your best attempt at a solution should go under 'your solution'.
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Assignment 12

@& If you substitute a for x^3, then you end up with a quadratic equation that can be easily factored. If a = x^3, then x^6 = a^2 so the equation becomes a^2 - 7 a - 8 = 0. We factor this and find that a can be either 8 or -1. So x^3 can be either 8 or -1. Thus x can be either 2 or -1. *@

@& The left-hand side consists of x^2 - 3x and the square root of x^2 - 3x. So instead of x^2 - 3 x - sqrt(x^2 - 3x) we write the left-hand side as u - sqrt(u), which is easier to deal with. We solve for u, then come back and figure out what value(s) of x give us our values of u. *@

@&

@& Check my notes and let me know if you have additional questions. Note that self-critiques and questions should go under self-critique, which is where I look for questions. Your best attempt at a solution should go under 'your solution'. *@

@&
If you substitute a for x^3, then you end up with a quadratic equation that can be easily factored.

If a = x^3, then x^6 = a^2 so the equation becomes

a^2 - 7 a - 8 = 0.

We factor this and find that a can be either 8 or -1.

So x^3 can be either 8 or -1.

Thus x can be either 2 or -1.
*@

@&
The left-hand side consists of

x^2 - 3x

and

the square root of x^2 - 3x.

So instead of

x^2 - 3 x - sqrt(x^2 - 3x)

we write the left-hand side as

u - sqrt(u),

which is easier to deal with.

We solve for u, then come back and figure out what value(s) of x give us our values of u.
*@

@&
Check my notes and let me know if you have additional questions.

Note that self-critiques and questions should go under self-critique, which is where I look for questions. Your best attempt at a solution should go under 'your solution'.
*@