Homework 14

#$&*

course Mth 158

10/23 8:34 pm

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:

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

First thing that I did is add the 1 to both sides to get the radical by itself.

This gave me:

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

Then I multiplied both sides by the radicand 3.

1-2x = -1

Then I added 1 to both sides.

-2x = 0

Then divide by -2, and that gave me

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): OK

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Self-critique Rating: 3

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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:

First this one was very confusing, I had to watch the MATH XL video on it several times to understand.

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

First I subtracted sqrt(x+2) from both sides.

Then I squared both sides to remove the radical.

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

which gives me:

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

Then I simplified

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

Then I subtracted x+3 from both sides

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

Then I squared both sides:

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

This gave me

(2x+4)(2x+4) = 4(x+2)

Then I used the distributive property

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

Then I subtracted 4x-8 from both sides

Which left me with

4x^2 + 12x + 8 = 0

Then I had to factor out the 4

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

Then I divided both sides by 4.

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

Then I worked (x+1) and (x+2) to get my solution set.

{-1,-2}

confidence rating #$&*: 2

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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. **

STUDENT QUESTION

I got to the third step but I got confused on what to eliminate or substitute in, looking at the solution, im still a

little confused on how it all worked out. U got any suggestions on how to look at it in a better way???

INSTRUCTOR RESPONSE

You're pretty much stuck with this technique and this way of looking at the problem.

It should be pretty clear to you that

(sqrt(x+3))^2 is just x + 3.

Squaring the expression [ -sqrt(x+2) +1] is a little more challenging.

We could use the distributive law:

[ -sqrt(x+2) +1]^2

= [-sqrt(x + 2) + 1 ] * [-sqrt(x + 2) + 1 ]

= -sqrt(x+2) * [-sqrt(x + 2) + 1 ] + 1 * [-sqrt(x + 2) + 1 ]

= -sqrt(x+2) * (-sqrt(x + 2) ) + (-sqrt(x + 2) * 1 + 1 * (-sqrt(x + 2) ) + 1 * 1

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

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

Once we get the equation

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

we see that we still need to 'get to' that x within the square root. To do that we rearrange the equation so that the

square root is on one side, all by itself, so we can square it without dragging a lot of other stuff along.

So we do a couple of steps and we get

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

If we square both sides of this equation, we get rid of all the square roots and we get x out where we can deal with it.

The details are in the given solution, but we get the equation

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

This equation now has x^2 and x terms, so we know it's a quadratic, and we rearrange and solve it as such. The details are

in the given solution.

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Self-critique (if necessary): OK

I do see where I should always check my work. Because I do see how -2 would be my only solution.

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Self-critique Rating: 2

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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:

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

First I factored out the x^(1/4)

This gave me:

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

Then I applied the zero principle to get

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

This gave me:

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

To complete this one I squared both sides of x^(1/2) = 9

This gave me x=0 and x=81 The solution set is {0,81}

confidence rating #$&*: 2

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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): OK

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Self-critique Rating: 3

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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:

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

First I let u=x^3 to solve this in quadratic form.

u^2 - 7u-8 = 0

(u-8)(u+1) = 0

u=8 an u=-1

Then I substituted u which is:

x^3=8 and x^3=-1

I solved these and came up with:

x=8^1/3 which equals 2.

x=-1^1/3 which equeals -1.

confidence rating #$&*: 2

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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.

STUDENT QUESTION

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)

INSTRUCTOR RESPONSE

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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Self-critique (if necessary): Ok

I used the letter u instead of a. I am assuming it is okay to use either variable.

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Self-critique Rating: 3

@&

The name of the variable is completely arbitrary, so you can use any unused symbol you wish.

*@

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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:

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

First I let u=sqrt(x^2 - 3x) to solve this in quadratic form.

u^2-u = 2

subtract 2 from both sides.

u^2 - u - 2 = 0

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

u = 2 and u = -1

Then I plugged the sqrt(x^2 - 3x) back in for u

I only checked for 2 because the solution can not be negative, so it can't be -1.

sqrt(x^2 - 3x) = 2

I squared both sides to remove the radical.

x^2 - 3x = 4

subtract 4 from both sides and then factored.

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

x=4 or x=-1

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

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.

STUDENT QUESTION

I got stuck on this part u=(-sqrt2+-sqrt10)/2, but after I looked at the solution, it made a little more sense to me, but

im not real confident. Got any suggestions on how to approach it in a different way???

INSTRUCTOR RESPONSE

Plugging into the quadratic formula we get

u=(-sqrt2+-sqrt10)/2,

meaning u can take one of the two values

u=(-sqrt2+sqrt10)/2

or

u=(-sqrt2-sqrt10)/2.

These quantities are just plain old numbers, which you could evaluate (up to some roundoff) on your calculator.

The first possible value of u is about equal to about .874.

The second possible value of u is negative.

Now u stands for x^2, so we ignore the negative value of u (this since x^2 can't be negative).

So we're left with

x^2 = u = .874.

So x = +- sqrt(.874), giving us the values of x in the given solution.

STUDENT QUESTION

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.

INSTRUCTOR RESPONSE

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.

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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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:

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

First I do understand the quadratic formula (finally!!) But I am very confused with this one. I see that I would let

u=x^2, but after that I am lost. I am not sure where or what to do with the square root. I looked at your solution but I

got confused where to go or how you came up with the 8 and how to solve. Any suggestions?

confidence rating #$&*: 1

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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

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.

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Self-critique (if necessary):

This one definitely confused me, I have really been taking my time with this chapter due to the fact that I am having a

harder time catching it. I have to read through the problems and work several before I understand these.

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Self-critique Rating: 1

@&

In the equation

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

the values of a, b and c are, respectively, 1, sqrt(2) and -2.

So when you substitute for b in the quadratic formula, you will substitute sqrt(2).

You'll get

x = (-sqrt(2) +- sqrt( (sqrt(2)^2 - 4 * 1 * (-2) ) ) / (2 * 1).

*@

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

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Self-critique (if necessary):

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Self-critique rating:

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

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Self-critique rating:

#*&!

&#Good responses. See my notes and let me know if you have questions. &#