OQ 14

#$&*

course Mth 158

6/24 1 pm

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 = 0+1

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

1-2x=1

1-2x-1=1-1

-2x / -2 =0/ -2

X=0

confidence rating #$&*: 2

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

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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

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

[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) + x+2

(2x+4) ^2 = [sqrt (x+2)]^ 2

(2x+4) (2x+4) = x+2

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

4x^2 +15x + 14= 0

(4x+14) (x+1) = 0

(4x+14) = 0 = -2

X+1 =0 = -1

confidence rating #$&*: 2

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

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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 did not check my solution and thus did not find that -1 was an extraneous solution.

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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.

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

u= x^(1/4) and u^2 = x ^2/4

u (u^2 -9) =0

u (u-3) (u-3) = 0

u=0 u-3=0

u= 0, -3

x= 0 x= -3 ^ (1/4) = 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: ok

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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

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

(x^3-8) = 0

X^3= 8

X= 2

(x^3 +1) =0

X^3 = -1= -1

confidence rating #$&*:

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

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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): 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.

[x^2 - 3 x - sqrt(x^2 - 3x)] ^2 = 2^ 2

x^2 - 3 x - x^2 - 3x= 4

-6x = 4

X= -2/3

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): I don’t understand why I could not preform the problem the way I did without substituting u for x…

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Self-critique Rating: 1

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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^2 = x^4

let u = sqrt(2) x^2

@& You can't use two different values for u. If u^2 = x^4, then it doesn't equal sqrt(2) x^2.*@

u^2 + u -2 = 0

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

sqrt(2) x^2 + 2 = 0

[sqrt(2)] ^ 2 x^2 + 2 = 0 ^2

2x^2 +2 =0

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

X= ˝, -1

sqrt(2) x^2 - 1 = 0

[sqrt(2)] ^ 2 x^2 - 1 = 0 ^2

2x^2 -1 =0

(2x- 1) (x-1) = 0

confidence rating #$&*: 2

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

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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 used the wrong terms that for u. I don’t understand why I couldn’t have used the formula I used.

"

Self-critique (if necessary):

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Self-critique rating:

@& Check my note, inserted into your solution, if you haven't already.*@

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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^2 = x^4

let u = sqrt(2) x^2

@& You can't use two different values for u. If u^2 = x^4, then it doesn't equal sqrt(2) x^2.*@

u^2 + u -2 = 0

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

sqrt(2) x^2 + 2 = 0

[sqrt(2)] ^ 2 x^2 + 2 = 0 ^2

2x^2 +2 =0

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

X= ˝, -1

sqrt(2) x^2 - 1 = 0

[sqrt(2)] ^ 2 x^2 - 1 = 0 ^2

2x^2 -1 =0

(2x- 1) (x-1) = 0

confidence rating #$&*: 2

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

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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 used the wrong terms that for u. I don’t understand why I couldn’t have used the formula I used.

"

Self-critique (if necessary):

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Self-critique rating:

@& Check my note, inserted into your solution, if you haven't already.*@

#*&!

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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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x^4 + sqrt(2) x^2 - 2 = 0

let u^2 = x^4

let u = sqrt(2) x^2

@& You can't use two different values for u. If u^2 = x^4, then it doesn't equal sqrt(2) x^2.*@

u^2 + u -2 = 0

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

sqrt(2) x^2 + 2 = 0

[sqrt(2)] ^ 2 x^2 + 2 = 0 ^2

2x^2 +2 =0

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

X= ˝, -1

sqrt(2) x^2 - 1 = 0

[sqrt(2)] ^ 2 x^2 - 1 = 0 ^2

2x^2 -1 =0

(2x- 1) (x-1) = 0

confidence rating #$&*: 2

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

.............................................

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 used the wrong terms that for u. I don’t understand why I couldn’t have used the formula I used.

"

Self-critique (if necessary):

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Self-critique rating:

@& Check my note, inserted into your solution, if you haven't already.*@

#*&!#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#