Assignment 12 14

#$&*

course Mth 158

11:00AM 2/22/13

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

Add 1 to both sides:

 (1-2x)^(1/3) = 1

Cube both sides:

 [(1-2x)^(1/3)]^3 = (1)^3

 1 - 2x = 1

 Subtract 1 from both sides:

 -2x = 0

 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: OK

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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:

 sqrt(3x+7) + sqrt(x+2) = 1

 Isolate sqrt(3x+7):

 sqrt(3x+7) = 1 - sqrt(x+2)

 Square both sides:

 (sqrt(3x+7))^2 = (1 - sqrt(x+2))^2

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

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

 3x + 7 = 1 - 2sqrt(x+2) + x + 2

 3x + 7 = 3 + x - 2sqrt(x+2)

 Add -3 and -x to both sides:

 2x + 4 = -2sqrt(x+2)

 Square both sides:

 (2x + 4)^2 = (-2sqrt(x+2))^2

 4x^2 + 16x + 16 = 4x + 8

 Add -4x and - 8 to both sides:

 4x^2 + 12x + 8 = 0

 Factor out 4:

 4(x^2 + 3x + 2) = 0

 4(x + 1)(x + 2) = 0

Zero product property:

 x = -1 or x = -2

 Check: x = -1

 sqrt(3(-1)+7) + sqrt(-1+2) = 1

 sqrt(4) + sqrt(1) = 1

 2 + 1 = 1

 3 ≠ 1

 Check x = -2

 sqrt(3(-2)+7) + sqrt(-2+2) = 1

 sqrt(1) + sqrt(0) = 1

 1 = 1

 Solution Set: {-2}

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. **

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

 

 

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Self-critique Rating: OK

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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

 Add 9 x^(1/4) to both sides:

 x^(3/4) = 9 x^(1/4)

 Raise each side by the power of 4:

 (x^(3/4))^4 = (9 x^(1/4))^4

 x^3 = 9x

@&

It took me a minute to catch this, so don't feel badly, but you didn't raise 9 to the 4th power.

*@

 divide both sides by x:

 x^2 = 9

 x = 3, or x = -3

 Check: x = 3

 3^(3/4) - 9 (3)^(1/4) = 0

 27^(1/4) - 27^(1/4) = 0

 0 = 0

 Check: x = -3

 (-3)^(3/4) - 9 (-3)^(1/4) = 0

 (-27)^(1/4) - (-27)^(1/4) = 0

@&

It was when I was checking your check that I noticed the error.

In the expression

9 (-3)^(1/4)

the 9 is not raised to the 1/4 power, only the -3.

In the expression

(-27)^(1/4)

the -27 represents 9 * -3, where the 9 was previously the coefficient of (-3)^(1/4). So now the 9 is part of an expression that is raised to the 1/4, whereas this was not formerly the case.

*@

 0 = 0

 Solution set: {3 , -3}

 

 

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):

 Okay, looking at the given solution, I can see it got its answer and was able to work it myself. Where did I go wrong with my work????

 

 

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Self-critique Rating: 2

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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

 u = x^3

 u^2 - 7u - 8 = 0

 (u - 8)(u + 1) = 0

 Zero product property:

 u = 8, or, u = -1

 u = x^3 = 8 = 2, or, u = x^3 = -1

 Check: x = 2

 (2)^6 - 7 (2)^3 - 8 = 0

 64 - 56 - 8 = 0

 0 = 0

 Check: x = -1

 (-1)^6 - 7 (-1)^3 - 8 = 0

 1 + 7 - 8 = 0

 0 = 0

 Solution Set: {2 , -1}

 

 

confidence rating #$&*: 3

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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

 

 

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Self-critique Rating: OK

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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

 Isolate - sqrt(x^2 - 3x):

 - sqrt(x^2 - 3x) = - x^2 - 3 x + 2

 Square both sides:

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

 x^2 - 3x = x^4 -3x^3 - 2x - 3x^3 + 9x^2 + 6x - 2x^2 + 6x + 4

 x^2 - 3x = x^4 - 6x^3 + 7x^2 + 10x + 4

 - x^4 + 6x^3 - 6x^2 - 13x - 4 = 0

 I'm not sure where to go from here. I tried factoring by grouping:

 (- x^4 + 6x^3) + (- 6x^2 - 13x - 4) = 0

 -x^3(x - 6) + (- 6x^2 - 13x - 4) = 0

 But I don't know where to take (- 6x^2 - 13x - 4).

 

 

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):

 Okay, I see how to get the answer and I worked it through it myself.

 

 

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Self-critique Rating: 2

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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'm not sure how to work this problem. I tried isolating the square root:

 x^4 + sqrt(2) x^2 - 2 = 0

 (sqrt(2))^2 = [(-x^4 + 2) / x^2]^2

2 = (x^8 + 4) / x^4

 And I tried solving in quadratic form:

 x^4 + sqrt(2) x^2 - 2 = 0

 u = x^2

 u^2 + sqrt(2)u - 2 = 0

 But am not sure how to factor that.

 

 

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):

 I don't understand how we go from:

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

to

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

Using the quadratic formula. Could you explain this please???? I understand how to get the solutions after u=(-sqrt2 +- sqrt(2-(-8))/2, but don't understand how to get there.

 

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Self-critique Rating: 1

@&

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

is of the form

a u^2 + b u + c = 0

with a = 1, b = sqrt(2) and c = -2.

Thus

b^2 - 4 a c = (sqrt(2))^2 - 4 * 1 * (-2) = 2 + 8 = 10

and the solution

u = (-b +- sqrt(b^2 - 4 a c) ) / (2 a) becomes

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

*@

 

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

"

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

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#*&!

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