Assign 12

#$&*

course Mth 158

6/23 0615am

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

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

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

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

1-2x=1

-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:

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:

(a+b)^2 = a^2 + 2 a b + b^2

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

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

confidence rating #$&*:

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

1

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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):

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???

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Self-critique Rating:

@&

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.

*@

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

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

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

x = 0 and x^(1/2) = 9

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:

a = x^3

a^2 - 7 a - 8 = 0

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

a = 8, a = -1.

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

x = 8^(1/3) = 2

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

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Self-critique (if necessary):

This question confused me a little, but I figured it out with a little help from the given solution.

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Self-critique Rating:

2

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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:

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

u^2 - u = 2.

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

u = 2, u = -1

sqrt(x^2 - 3 x) = 2

sqrt(x^2 - 3x) = 2

x^2 - 3x = 4

x^2 - 3x - 4 = 0

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

x = 4 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.

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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

u=x^2 u^2 = x^4

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

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

u=(-sqrt2+-sqrt10)/2

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 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???

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Self-critique Rating:

@&

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.

*@

2

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:

2

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:

#*&!

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