course Mth 158 ¡éþwWŸýDëÆØæ³É|ôÁ¤‹Ó½z¥í•ÆÖassignment #012
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15:12:15 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 --> Cube both sides [(1-2x)^(1/3) - 1]^3 = 0 ^3 1-2x -1 = 0 +1 +1 = 1-2x = 1 -1 -1 = -2x = 0 divide both sides by -2 x = 0 confidence assessment: 2
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15:12:25 ** 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 self critique assessment: 3
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15:48:12 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 --> move one radical over to the other side sqrt(3x+7) + sqrt(x+2) = 1 - sqrt(x+2) - sqrt(x+2) sqrt(3x +7) = 1 - sqrt (x+2) square both sides [sqrt(3x +7)]^2 = [1 - sqrt (x+2)]^2 3x+7 = 1 - sqrt (x-2) - sqrt (x-2) + x + 2 -x -3 -1 - x - 2 = 2x + 4 = -2 sqrt (x+2) square both sides again [2x + 4]^2 = [-2 sqrt (x+2)]^2 4x^2 + 16x +16 = 4 + x + 2 -x -2-4 = 4x^2 + 15x + 10 = 0 Since I couldn't figure out how to factor it correctly, I used the quadratic formula. x = -15 +/- sqrt (15^2 - 4 * 4 * 10) / 2 * 4 x = -15 +/- sqrt (225 - 160) / 8 x = -15 +/- sqrt (65) / 8 x = -15 +/- 8 sqrt (1) / 8 x = -15 + sqrt(1) or x = -15 - sqrt (1) Simplifying more... x + 15 = sqrt(1) which we square both sides again (x + 15) ^2 = [sqrt(1)]^2 x^2 + 30x + 225 = 1 x^2 + 30x + 224 = 0 (x+14)(x+16) = 0 So x = -14 or -16 {-14, -16} confidence assessment: 2
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15:53:03 ** 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 see where I went wrong - at the point where I got to 4x^2+16x+16= 4(x+2) I wrote instead 4x^2+16x+16= 4+x+2. Simple mistake, bad outcome! self critique assessment: 2
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16:00:08 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 --> move one over to the other side x^(3/4) = 9 x^(1/4) Multiply each side times the 4th power [ x^(3/4)] ^4 = [9 x^(1/4)]^4 x^3 = 9x
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16:01:05 ** 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 --> not understood self critique assessment: 1
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16:10:02 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 --> Factoring we get (x^3 - 8)(x^3 + 1) = 0 setting (x^3 - 8) to 0 we get x=2 setting 9x^3 + 10 to 0 we get x = 1 x = 2 is the only one that works when checking the problem. x = 2 confidence assessment: 2
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16:10:30 ** 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 self critique assessment: 2
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16:32:37 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 --> Square both sides to get rid of the sqrt [ x^2 - 3 x - sqrt(x^2 - 3x)]^2 = 2^2 (x^2 - 3x)(x^2 - 3x) + x^2 - 3x = 4 x^4 - 6x^3 + x^2 + 6x = 4 x(x^3 - 6x^2 + x + 6) = 4 I don't know what to do from here, because when I try to group the 4 problems, it doesn't turn out. confidence assessment: 1
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16:33:42 ** 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 see that I forgot to separate the sqrt first before anything else. self critique assessment: 2
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16:54:51 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 --> Single out the sqrt x^4 - 2 = - sqrt(2) x^2 (x^4 - 2)^2 = (-sqrt(2) x^2)^2 x^8 - 4x^4 + 4 = 2x^4 - 2x^4 -2x^4 = x^8 - 6x^4 + 4 = 0 x = 6 +/- sqrt(6^2 - 4*1*4) / 2*1 x = 6 +/- sqrt(36 - 16) / 2 x = 6 +/- sqrt (20) / 2 x = 6 +/- 4 sqrt (4) / 2 x = 6 +/- (4 * 2) / 2 x = 6 + 4, x = 6-4 x = 10, x = 2 {10, 2} confidence assessment: 1
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16:56:22 ** 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 --> Understood. I didn't realize how to properly use the ""u"" method. self critique assessment: 2
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16:56:28 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> confidence assessment:
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