Assignment 12

course Mth 158

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11:52:30

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

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

1 - 2x = 1

2x = 0

x = 0

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11:52:35

** 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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11:59:26

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

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

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

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

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

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

7x^2 + 12x + 8 = 0

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

x + 1 = 0 so x = -1

x + 2 = 0 so x = -2

{-2}

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11:59:56

** 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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12:03:33

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

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

x^3 = 6561x

x (x^2 - 6561)

x (x + 81)(x + 81) = 0

x = 0, -81, 81

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12:03:59

** 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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12:05:46

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

x^3 - 8 = 0

x^3 = 8

x = 2

x^3 + 1 = 0

x^3 = -1

x = -1

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12:06:07

** 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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12:13:17

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

au^2 + bu + c = 0

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

u^2 - u = 2

u^2 - u - 2 = 0

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

u - 2 = 0

u = 2

u + 1 = 0

u = -1

{2,-1}

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12:13:31

** 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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12:17:32

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

Im not sure where to start. Help

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12:19:40

** 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, How do you determine to use the quadratic formula?

It's something you learn to be on the lookout for, because you have a relatively simple formula for solving it.

When you have different powers of the variable, you need to identify what you have. With different powers it's not linear. So the first question you ask is if it could be expressed as a quadratic, which is the next-easiest type of equation to solve. If not, you see whether it might be a polynomial, which is generally less easy to solve unless you can somehow factor it.

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12:20:05

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

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none

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Assignment 12

Good. See my note and let me know if you have questions.