Assign 12

course Mth 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:

(1-2x)^(1/3) - 1 = 0 First you add 1 to both sides.

(1-2x)^(1/3) = 1 Then raise each side of the equation by 3

[(1-2x)^(1/3)]^3 = 1^3 This changes the equation to

(1-2x)^(1/3) * (3/1) = 1*1*1 since 1/3 * 3/1 = 3/3 = 1 you have

(1-2x) = 1 adding -1 to both sides

1-1-2x = 1-1 leaving us with

-2x = 0 divide each side by -2

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):

There were no discrepancies.

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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:

sqrt(3x+7) + sqrt(x+2) = 1 first isolate the complex on the left side

sqrt(3x+7) = -sqrt(x+2) +1 then square both sides

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

3x+7 = x+2 –sqrt(x+2) +1 isolate the –sqrt on the right

3x-x+7-3 = -sqrt9x+2) add like terms

2x + 4 = -sqrt(x+2) square both sides

(2x+4)^2 = (-sqrt(x+2)^2 distribute on the right side

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

4x^2 + 16x + 16 = 4x + 8 use zero product by adding -4x-8 to both sides

4x^2 + 16x -4x +16 -8 = 0 then factor

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

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

x=-1 or x=-2 when checking, only one works so

{-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. **

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Self-critique (if necessary):

There were no discrepancies

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Self-critique Rating:3

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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 I wasn’t sure what the (3/4) was. I tried to write the given solution down on paper and follow it, but could not do it. I will take this one to my tutor this evening.

(3/4) is preceded by ^, so that makes it a power.

An alternative to the solution given below:

Add 9 x^(1/4) to both sides to get

x^(3/4) = 9 x^(1/4). Raise both sides to the power 4:

(x^(3/4)^(4)) = (9 x^(1/4))^4. Use the laws of exponents to get

x^(3/4 * 4) = 9^4 * x^(1/4 * 4). Simplify (no need to calculate 9^4 at this point; I think that comes out 6641 but it doesn't matter) to get

x^3 = 9^4 * x^1. Divide by x^1 to get

x^2 = 9^4. Take the 1/2 power of both sides to get

x = 9^(4)^(*1/2) = 9^(4 * 1/2) = 9^2 = 81.

confidence rating: 0

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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): 0

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Self-critique Rating: 0

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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 If we substitute for the x^6 and let u = x^3 and u^2 = x^6 we can re- write the equation as follows:

u^2 – 7u -8 = 0 now it can be factored as a quadratic equation

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

u=8 or u=-1 now that we have the equation for u, we can then solve for x

if u = x^3=8 or u=x^3=-1 then

x=8^(1/3) = 2

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

{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.

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Self-critique (if necessary):

There were no discrepancies

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Self-critique Rating:3

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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 let u = sqrt(x^2-3x) and u^2 = x^2-3x

u^2 – u = 2 zero factor by adding -2 to both sides

u^2 – u -2 =0 factor

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

u =2 or u =-1 from here solve for x

we know the square of a real number can not be a negative number

if sqrt of x^2-3x =2 then

x^2 -3x =4 add -4 to both sides

x^2 -3x -4 =0

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

x =4 x =-1

{-1, 4}

confidence rating: 3

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

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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):

There were no discrepancies

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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 let u = x^2 and u^2 = x^4

u^2 + sqrt(2) I got lost here. I couldn’t figure out if the next step was to multiply u-2? the way it is in the original sqrt(2) x^2 doesn’t make sense to me.

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

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 am still confused. I will take this one to my tutor also.

You nearly had the equation. The quadratic formula and its interpretation add a bit of challenge, of course.