course Mth 158 ^ύƯwƅқassignment #013
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11:29:44 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 added 1 to both sides (1-2x)^1/3 = 1. Then I cubed both sides. 1 - 2x = 1. Next I subtracted 1 from both sides -2x = 0 divide by -2 and I got x = 0. confidence assessment: 2
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11:30:22 ** 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 --> That is how I did it. self critique assessment: 3
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11:32:43 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 --> First I squared both sides 3x+7 + x+2 = 1. Then I combined like terms 4x + 9 = 1. Then subtracted 9 from both sides. 4x = -8 and divided by 4 to get x = -2. confidence assessment: 2
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11:33:59 ** 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 just squared both sides and got -2. I see how the other way would have worked too. self critique assessment: 2
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11:39:53 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 --> Since both parts of the equation take the fourth root, I raised both sides of the equation to the 4th power and got x^3 - 9x = 0. Then I factored out x and got x(x^2 - 9) = 0. I set x = 0 and x^2 - 9 = 0. I then solved for x by adding 9 to both sides x^2 = 9 took the square root and got x = 3. So my answers are x = 0, 3. confidence assessment: 1
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11:42:46 ** 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 zero as an answer, but I didn't get 81. I see how x^(1/4) factors out and how you got 81 as a solution. self critique assessment: 2
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12:04:40 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 --> I factored x^6 - 7x^3 - 8 = 0 into (x^3 - 8) ( x^3 + 1) = 0. then I set x^3 - 8 = 0 and added 8 to both sides x^3= 8 took the cube root and found x = 2. Then I set x^3 + 1 = 0 subtracted 1 from both sides and took the cube root and got x = 1. confidence assessment: 2
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12:05:04 ** 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 --> I got it. self critique assessment: 3
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12:16:21 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 --> I squared both sides to get x^4 - 3x^2 - x^2 - 3x = 4. Then I combined like terms and got x^4 - 4x^2 - 3x = 4. I factored out a x and got x(x^3 - 4x - 3) = 4. I set x=4 and x^3 - 4x -3=4. I factored that to (x^3 - 3) (x-1). x^3 - 3 = 4 and got x = cube rt (7). and x -1 = 4 and found x= 5. Since cube rt(7) is not a real number, the solutions for this problem are ( 4 , 5) confidence assessment: 1
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12:17:53 ** 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 got that one wrong. I see how u can equal x^2 - 3x. I understand how to rearrange the problem and get the correct solutions. self critique assessment: 2
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12:29:02 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 --> Let u = x^2 so u^2 = x^4. u^2 + sqrt (2u) - 2 = 0. Then I add 2 to both sides u^2 + sqrt (2u) = 2. Then I squared both sides and got u^4 + 2u = 4. I factored out an u and found u(u^3 + 2). I set u=4 and u^3 + 2=4. subtracted 2 from both sides and got u = cube rt(2) which rounded equals 1.26. So my solutions are 4 and 1.26. confidence assessment: 1
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12:30:37 ** 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 didn't use the quadratic formula. I know the formula and how to use it. I understand how it can be used to get the correct solutions. self critique assessment: 2
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12:31:16 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I learned how the quadratic formula can be used to find the solutions to these problems. confidence assessment: 3
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