Assignment 12

course mth 158

ǥGظm䇎n~{assignment #012

012. `query 12

College Algebra

07-10-2008

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02:06:25

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 -->

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

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

1-2x=1

-2x=0

x=0

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02:06:36

** 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 -->

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assignment #012

012. `query 12

College Algebra

07-10-2008

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13:20:43

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 -->

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

(1-2x) = 1

2x+1=1

2x=0

x=0

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13:20:47

** 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 -->

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13:24:49

**** query 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 -->

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

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

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

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

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

4x^2+12x+12

4(x^2+3x+3)

4(x+2)(x+1)

x= -2 or x= -1

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13:25:11

** 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 -->

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13:28:53

**** query 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 -->

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

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

x^3=9x

x(x^2)=x(9)

x^2=9

x=3

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13:31:12

** 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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RESPONSE -->

ok i see

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^(1/2) = 9

x = 81 or x=0

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20:06:33

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

x^3=a

ax^3 - 7a -8 =0

ax^3-7a=8

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20:08:58

** 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 is see my mistake

a=x^3

so

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

a^2 - 7 a - 8 = 0

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

a = 8 or a = -1.

x^3=8 or x^3= -1

x=2 or x= -1

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20:21:45

**** query 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 -->

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

x^4- 3 x^3 -(x^2 - 3x) = 4

x^2((x^2 - 3x) - (x^2 - 3x)=4

(x^2 - 3x)=a

ax^2-a=4

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20:26:23

** 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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RESPONSE -->

ok i see what i did i needed to take x^2-3 out for a and i only took x^2 and i also should have left it sqrt

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

u^2 - u = 2.

u^2 - u - 2 = 0

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

u = 2 or u = -1.

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20:58:49

**** query 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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RESPONSE -->

x^2=a

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

a^4+ 2a^2 -4=0

(a^2+2a-2)^2=0

a^2+2a-2=0

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21:03:26

** 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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RESPONSE -->

ok i really dont think I get this problem. Where did all the sqrts come from??

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21:03:34

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

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Your work looks very good. Let me know if you have any questions. &#