* 1.4.12 (was 1.4.6). Explain how you found the real solutions of the equation (1-2x)^(1/3) - 1 = 0 ********************************************* Your solution: Take the 1 to the other side and exponent of three to both sides giving [(1-2x)^(1/3)]^3 = 1^3 because of law of exponents the exponent cancels out and we get 1-2x=1 and now we solve so we solve through and x=0 &# Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Had to think twice about if zero will work like in the other assignment we had x=0 but when you plug it back into the denominator it equaled 0 which cannot happen Self-critique Rating: ********************************************* Question: * 1.4.28 (was 1.4.18). Explain how you found the real solutions of the equation sqrt(3x+7) + sqrt(x+2) = 1. ********************************************* Your solution: I’m a bit confused because we need to get rid of the exponent however multiplying by ½ may not make sense because then on the right hand side we would have 1^1/2.. Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. This 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 As it turns out: the solution -1 gives us sqrt(4) + sqrt(1) = 1 or 2 + 1 = 1, which isn't true, while the solution -2 gives us sqrr(1) + sqrt(0) = 1 or 1 + 0 = 1, which is true. • x = -1 is an extraneous solution that was introduced in our squaring step. • Thus our only solution is x = -2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Okay so you must use the expanding rule/formula and then use trial and error to see which answer fits in. I didn’t recognize this when I saw the problem this time. Self-critique Rating: ********************************************* Question: * 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. ********************************************* Your solution: Take out ¼ from the exponent and get x^(1/4) ( x^(1/2) - 9) = 0 and because =0 then solve each x expression to = so x=0 and x=81 Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: * 1.4.46 (was 1.4.36). Explain how you found the real solutions of the equation x^6 - 7 x^3 - 8 =0 ********************************************* Your solution: Looking ahead, I didn’t realize to make a=x^3 and then a^2 = x^6 and now substituting the a in place of x and then factoring which gives you a=8 and a=-1 and solve with these answers in a = x^3 giving 2 and -1 Confidence Assessment: I had to think twice a little bit on this one in how to approach the problem ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. ********************************************* Question: * 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. ********************************************* Your solution: Because of the ‘real solutions’ part of the question like above we use this substitution of a= sqrt(x^2 - 3x) so a^2= x^2 - 3x so a^2 - a - 2 = 0 and factoring to get (a-2)(a+1) = 0 and solving gives a = 2 and a = -1 then substitute x^2 - 3x and we get sqrt(x^2 - 3 x) = 2 and sqrt(x^2 - 3 x) = -1 so x^2 - 3x = 4 then x^2 - 3x - 4 = 0 factor and get (x-4)(x+1) = 0 and finally x = -4 or x = 1. The other expression cannot happen because square root can not be negative Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ this was easier once I went through the previous questions and understood it better
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Given Solution: * * 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There is a lot of substituting so you have to be careful at which time to substitute and calculate properly Self-critique Rating: ********************************************* Question: * 1.4.92 \ 90 (was 1.4.66). Explain how you found the real solutions of the equation x^4 + sqrt(2) x^2 - 2 = 0. ********************************************* Your solution: a = x^2 so a^2=x^4 so a^2 + sqrt.(2) a-2 then this one is a bit confusing now not sure how to go forward. But looking forward I see that using the quadratic formula now it what to do. Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Starting with x^4+ sqrt(2)x^2-2=0 we let u=x^2 so that u^2 = x^4 giving us the equation u^2 + sqrt(2)u-2=0 Using the 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 the equation come from 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 to three significant figures are • x = .935 and • x = -.935. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): You must remember the previous laws and formulas and apply then now. Self-critique Rating: Add comments on any surprises or insights you experienced as a result of this assignment. This assignment was a good one because I knew some things but couldn’t follow through on all of them, so I learned a lot.