Assignment 12

course MTH 158

|~RԷvcӞŏzzassignment #013

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013. `query 13

College Algebra

09-29-2007

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15:16:03

query 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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RESPONSE -->

first, get radical by itself:

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

raise each side to third power:

1-2x=1

-2x=0

x=0

confidence assessment: 3

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15:16:11

** 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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RESPONSE -->

ok

self critique assessment: 3

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15:27:42

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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RESPONSE -->

first isolate one radical:

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

square each side:

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

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

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

square each side again

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

4x^2+12x+8=0

4x(x+1) +8(x+1)

(x+1)(4x+8)

x+1=0 x=-1

4x+8=0 4x=-8 x=-2

{-1,-2}

confidence assessment: 2

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15:27:55

** 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, which 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

It turns out that the -1 gives us sqrt(4) + sqrt(1) = 1 or 2 + 1 = 1, which isn't true, while -2 gives us sqrr(1) + sqrt(0) = 1 or 1 + 0 = 1, which is true.

x = -1 is the extraneous solution that was introduced in our squaring step.

Thus our only solution is x = -2. **

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RESPONSE -->

ok

self critique assessment: 3

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15:34:08

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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RESPONSE -->

I am not sure

confidence assessment: 0

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15:34:24

09-29-2007 15:34:24

** 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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NOTES ------->

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Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the given solution, and if necessary asking specific questions (to which I will respond).

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15:37:32

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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RESPONSE -->

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

x^3-8=0 x^3=8 x=2

x^3+1=0 x^3=1 x=1

{-2,1}

confidence assessment: 2

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15:37:52

** 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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RESPONSE -->

ok

self critique assessment: 3

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15:44:15

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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RESPONSE -->

Let u represent sqrt(x^2-3x) and u^2 be x^2-3x

u^2-u=2

u^2-u-2=0

u^2 -2u+1u-2=0

u(u-2) 1(u-2)

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

u=-1;u=2

confidence assessment: 1

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15:44:19

09-29-2007 15:44:19

** 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.

DER **

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NOTES ------->

Good start, but self-critique might in order for the second part. Chances are, however, that you understand it.

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15:46:34

1.4.90 (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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RESPONSE -->

i am not sure

confidence assessment: 0

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15:46:39

09-29-2007 15:46:39

** Starting with

x^4+ sqrt(2)x^2-2=0 we let u=x^2 so that u^2 = x^4:

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

using 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

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 are x = .935 and x = -.935. **

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NOTES ------->

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15:47:15

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

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RESPONSE -->

the last two questions I struggled with, I wasn't even sure where to start.

confidence assessment: 3

You had a good start on #64.

Do you understand the solution to #90?

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Overall I think you're doing OK here.

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Let me know if you have questions. &#