Assignment12

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22:19:55 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 --> i first set whats in () to the 3rd sqrt because of the facton 1/3 3rd sprt(1-2x)-1=0 now add the one to the other side 3rd sqrt(1-2x)=1 now set each side to the 3rd power (1-2x)^3=1^3 you get 1-2x=1 subtract the 1 to the other side -2x=0 x=0 Not sure about the answer

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22:20:06 ** 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 --> ok

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22:24:03 **** 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 --> i multiplied everything by the 2nd power (sqrt(3x+7)^2 + (sqrt(x+2)^2=1^2 so now you have 3x+7 + x+2=1 add the like terms 4x+9=1 subtract the nine 4x=-8 divide by 4 to get x=-2

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22:24:16 ** 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 --> ok

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22:26:24 **** 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 --> there no are real solutions

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22:28:21 ** 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

Your response did not agree with the given, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the given solution, and if necessary asking specific questions.

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22:35:19 **** 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 --> you need to factor out a x^3 because its whats in common between the grouped x^6-7x but as far as the common factor i'm lost

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22:37:40 ** 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

You need a detailed self-critique here.

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22:39:41 02-08-2006 22:39:41 **** 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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NOTES ------->

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22:39:45 **** 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 --> ok

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22:39:54 ** 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

This also requires a self-critique.

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22:41:48 **** 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 --> i don't know

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22:41:49 **** 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 -->

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22:42:35 ** 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 don't understand setting the problem up to other letters

This problem involves x^4 and x^2. You need to recognize that both are powers of x^2, since x^4 = (x^2)^2. Then you can let u = x^2 so that x^4 = u^2 and x^2 = u.

This reduces the equation to a quadratic, involving just u^2 and u. Once you have the equation in the form of a quadratic, you can solve it by factoring or, if that doesn't work (as is often the case), by the use of the quadratic formula.

The solution you get is for u, not x. Since u = x^2, you can just take +- sqrt(u) to get x.

A previous problem involved x^6 and x^3. Once again you have to recognize these as both being powers of x^3, with x^6 = (x^3)^2. Letting u = x^2, you could also write this equation as quadratic in u, and proceed from there.

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22:42:52 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE --> I don't understand the last few....i'll read up in the chapter

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You did well on some of these problems, but should have included more detail about your solution and your thinking on most problems, and most should have been self-critiqued but weren't. See my notes, in boldface like this one, to be sure you understand what is mean by 'self-critique', and when you need to do it. I only marked the first couple of places where you should have self-critiqued, but most of your responses required it.

Self-critique, when it is needed, is the most important part of the Query process. It gets you to focus on important details, and it tells me what I need to know in order to help.