Assignment 12

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09:17:57 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)^1/3] ^3 = 1^3 1-2x ^1 = 1 -2x = 0 x = 0

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09:21:05 ** 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 --> correct

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09:49:25 **** 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 isolate more complicated radical sqrt(3x + 7) = -sqrt(x + 2) square both sides (sqrt(3x + 7))^2 = ((-sqrt(x+2)^2) + 1)^2 remove parentheses (3x + 7) = x + 2 - 2(sqrt x+2) + 1 simplify and add like terms 3x + 7 = x + 3 - 2sqrt(x+2) isolate remaining radical 2x + 4 = -2sqrt(x + 2) square both sides (2x + 4)^2 = (-2sqrt(x+2))^2 remove parentheses 4x^2 + 16x + 16 = 4 (x + 2) distribute right side 4x^2 + 16x + 16 = 4x + 8 leave zero on one side 4x^2 + 12x + 8 =0 factor (2x + 2)(2x + 4) = 0 get solution 2x +2 = 0 ; 2x + 4 = 0 x = -1 ; x = -2 {-1,-2}

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09:57:40 ** 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 --> I got it correct but I figured I got it wrong. I just factored different that you did at the end of the problem.

You got a good solution. Because the result is easier to factor, it is good practice to factor out the largest possible monomial before factoring into binomials.

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09:57:17 **** 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 multiply 4 to each(since the denominators of the powers are 4) (x^(3/4))^4 - (9x^(1/4))^4 = 0^4 simplify x^3 - 9x^1 = 0 factor x(x^2 -9) = 0 x(x+3)(x-3) = 0 x+3=0 ; x-3=0 x = -3 ; x = 3 {-3,3}

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09:57:44 ** 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 --> I got this one wrong. I don't know exactly how it became ^1/4 and ^1/2 though.

x^(1/4) is a monomial common to both terms.

Be sure you see first why x^(1/4) ( x^(1/2) - 9) = x^(3/4) - 9 x^(1/4) (just multiply it out).

Then you should be able to see how it works in reverse.

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10:06:00 **** 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 u^2 - 7u - 8 = 0 (u-8)(u+1) = 0 u = 8 ; u = -1 After this, I really don't know what to do. I used example 4 in the book guide me. But I still don't understand how to do that at all. :(

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10:17:26 ** 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 --> Even after reviewing what you did, I don't understand how you got ^1/3 out of ^3. ??? Sorry.

To solve

x^3 = 8 we take the 1/3 power of both sides to get (x^3)^(1/3) = 8^(1/3). By the laws of exponents (x^3)^(1/3) = x^(3 * 1/3) = x^1 = x, so we have x = 8^(1/3).

8^(1/3) is the number which when multiplied by itself 3 times gives you 8--i.e., 8^(1/3) is the cube root of 8, or 2.

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10:40:02 **** 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 let p = sqrt(x^2 - 3x) => p^2 = x^2 - 3x p^2 - p = 2 p^2 - p - 2 = 0 (p-2)(p+1) = 0 p = 2 p = -1 -1 won't work because it has to be a non-negative. p = 2 = sqrt(x^2 - 3x) = 2 sqrt(x^2 - 3x) = 2 x^2 - 3x = 4 x^2 - 3x - 4 = 0 (x-4)(x+1)=0 x = 4 or x = -1

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10:45:24 ** 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 --> I am getting it now. I used number 63 to help me. I used the book maroon book that came with my Mth 158 books that has the odd problems worked out. Now I know how to do these. They are hard to get started. But, after you get started, they are easy to do.

Hard to get started, easy enough once you get started. That's pretty much the story of most basic mathematics.

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10:50:30 **** 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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RESPONSE --> x^4 + sqrt(2) x^2 - 2 = 0 I have no clue how to solve this one or where to even start. I tried like I did last time with the maroon book, but I don't understand that at all.

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10:51:13 ** 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 --> I am still confused.

You need to tell me exactly what you do and do not understand about this solution so I can help you through the things you don't understand.

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10:51:50 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE --> This one was pretty hard. It started out easy, but it got more complicated as it went on.

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Good work. Let me know if you have questions.