assignment 12

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course Mth 158

10/4 11

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:

start with (1-2x)^(1/3) - 1 = 0 add 1 to both sides

(1-2x)^(1/3)= 1 cube each side so you have

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

1-2x= 1 subtract 1 from both sides

-2x=0 divide both sides by zero, but you get 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: 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)+ 1

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

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

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

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

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

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

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

4x^2+12x+8=0

4(x^2+3x+2)=0

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

x+1=0 or x+2=0

x=-1 or x=-2

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

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

1+0=1

-2 checks out

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

2+1=1

-1 does not check out

so x=-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): ok

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Self-critique Rating:ok

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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:

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):

I don't get how you factored x^(1/4) from both sides. I get the zero principle and all the steps after it, but I don't understand how you got from x^(3/4) - 9 x^(1/4) = 0 to x^(1/4) ( x^(1/2) - 9) = 0. My brain just isn't working with me on this one.

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Self-critique Rating:ok

@& Start by multiplying out the expression x^(1/4) ( x^(1/2) - 9).

Now, to factor, say, x^2 out of x^4 + 3 x^3 + 5 x^2 what do you do? There are a lot of ways to look at this but ultimately what you are doing is dividing x^4 + 3 x^3 + 5 x by x^2, getting the result x^2 + 3 x + 5. Since this is what you get when you divide the original expression by x^2, it follows that x^2 multiplied by your result is equal to the original expression. That is, x^2 ( x^2 + 3 x + 5) = x^4 + 3 x^3 + 5 x^2.

Similarly, to factor x^(1/4) out of the expression in this problem, you divide that expression by x^(1/4), etc..*@

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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

fine, a =x^3 so

a^2- 7x-8=0

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

a=8 or a=-1

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

x=2 or x=-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):

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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. let u^2= (x^2-3x) then you have

u^2-u = 2.

u^2-u-2=0

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

u-2=0 or u+1=0

u=2 or u= -1

However, as said earlier u^2= (x^2-3x) so we'd have to plug it back in as

2^2=x^2-3x or -1^2=x^2-3x so solve that to get

4=x^2-3x or 1=x^2-3x

x^2-3x-4=0 or x^2-3x-1=0

(x-4)(x+3)=0 but the other equation doesn't work out because no sum of product of -1 can equal -3

so x-4= 0 or x+3=0

x=4 or x=-3

4 checks out but -3 does not

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

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 guess I did it a little differently, but does my way still work?

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Self-critique Rating:ok

@& Letting u = sqrt(x^2 - 3 x) is pretty much the same as letting u^2 = x^2 - 3 x.

However there is a difference. If u = sqrt(x^3 - 3 x) then u can't be negative.

So we can dismiss the u = -1 solution immediately.

However anything resulting from the u = -1 solution is going to be dismissed eventually, so it does no harm to start with u^2 = x^2 - 3 x.

You got x = -3, but that comes from an incorrect factoring of the quadratic (you get (x - 4) ( x + 1), not (x - 4) ( x + 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 so you have

u^2+sqrt(2)u-2=0 then use the quadratic formula to solve so

a=1 b=sqrt(2) and c=-2

[-sqrt(2) (+-) sqrt(sqrt(2)^2 -4*1*-2)]/2*1

[-sqrt(2) (+-) sqrt(2+8)]/2

[-sqrt(2) (+-) sqrt(10)]/2

from here I used the calculator and got u=.87 or u=-2.29

once again we said that u=x^2 so basically you just have to find the sqrt(u) to find x and you get (when using the calculator)

x=.93 but the -2.29 won't work because it's negative anyway, so really you're just left with x=.93

confidence rating #$&*: 3-ish

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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

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):ok

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Self-critique Rating:ok

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assignment 12

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