Chap 14 assignment

#$&*

course Mth 158

07/20 11:32pm

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

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

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

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

1-2x=1

2x=1-1

2x=0

X=0

confidence rating #$&* 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I got through this problem be referring to the text on almost every step

------------------------------------------------

Self-critique rating #$&* Fair

*********************************************

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

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

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

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

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

(2x+4)^2=(-2sqrtx+2)^2

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

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

4x^2+12x+16=8

4x^2+12x+16-8=8-8

4x^2+12x+8=0

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

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

{-1, -2}

confidence rating #$&* 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I am confused in the solution which I followed to get to the end. the step (sqrt3x+7)^2=[-sqrt(2+x)+1]^2 =3x+7=x+2-2sqrt(x+2)+1 I don’t understand why that become –x+2+1.

I don't think there's anything in the given solution that looks like –x+2+1. Can you clarify your question? You could copy this problem into the Question Form and clarify exactly what you mean by –x+2+1?

------------------------------------------------

Self-critique rating #$&*????

*********************************************

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

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

X^3=9^4*x

X(x^2+9^4)=0

should be X(x^2-9^4)=0

X(x-9^2)(x+9^2)=0

X=0

X=81

confidence rating #$&* 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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). **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

My solution looks different from the given solution.

Your solution is a valid alternate path through the problem.

------------------------------------------------

Self-critique rating #$&*Fair

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

X^6-7x^3-8=0

(x^3-8)(x^3+1=0

X^3-8+8=0+8

X^3=8

X=2

X^3+1-1=0-1

X^3=-1

X=-1

confidence rating #$&* 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating #$&* Fair

*********************************************

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

U=sqrt(x^2-3x)

U^2=x^-3x

u^2-u+2=0

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

U=2 u=-1

When reading the solution, I got lost at this point. I don’t understand the next step.

confidence rating #$&* 0

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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 sqrt(x^2 - 3x) back in for u we get

note the sqrt wasn't in this line before; this might have been your point of confusion

we started by letting U=sqrt(x^2-3x), as in your solution.

We solve the resulting equation for u.

Now we are substituting sqrt(x^2-3x) for u, which since u = 2 or u = -1 gives us the following two equations:

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

so

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

See question in my solution.

------------------------------------------------

Self-critique rating #$&*

*********************************************

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

U=x^2

U^2=x^4

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

U=-sqrt2+sqrt2-(-8)/2

U=-sqrt2+sqrt10/2

U=x^2

X^2=sqrt(sqrt-2+10)/2

X^2=sqrt(1.41-3.16)/2

X^2=sqrt(1.75)/2

X^2=.875

X=.9354

confidence rating #$&* 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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

so

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I am still reading the text and checking the solution with each step I do.

------------------------------------------------

Self-critique rating #$&*

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

I found this assignment to be very hard to follow from step to step in the equations. I am writing them out using symbols and then I get confused when I start typing.

Typing the expressions makes you think about what they mean, and you're getting pretty good at that. You sometimes leave out a sign of grouping, but I can follow your thinking.

How are you doing with the problem assignments as a whole? Are you able to get answers on the odd-numbered problems that match those in the back of the book?