course Mth 158 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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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): There were no discrepancies. ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: sqrt(3x+7) + sqrt(x+2) = 1 first isolate the complex on the left side sqrt(3x+7) = -sqrt(x+2) +1 then square both sides (sqrt(3x+7)^2 = (-sqrt(x+2) +1)^2 simplify 3x+7 = x+2 –sqrt(x+2) +1 isolate the –sqrt on the right 3x-x+7-3 = -sqrt9x+2) add like terms 2x + 4 = -sqrt(x+2) square both sides (2x+4)^2 = (-sqrt(x+2)^2 distribute on the right side 4x^2 + 16x + 16 = 4(x+2) 4x^2 + 16x + 16 = 4x + 8 use zero product by adding -4x-8 to both sides 4x^2 + 16x -4x +16 -8 = 0 then factor 4((x+1)(x+2)) = 0 (x+1)(x+2) = 0 x=-1 or x=-2 when checking, only one works so {-2} confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): There were no discrepancies ------------------------------------------------ Self-critique Rating:3 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^(3/4) - 9 x^(1/4) = 0 I wasn’t sure what the (3/4) was. I tried to write the given solution down on paper and follow it, but could not do it. I will take this one to my tutor this evening.
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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): 0 ------------------------------------------------ Self-critique Rating: 0 ********************************************* 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^6 - 7 x^3 - 8 =0 If we substitute for the x^6 and let u = x^3 and u^2 = x^6 we can re- write the equation as follows: u^2 – 7u -8 = 0 now it can be factored as a quadratic equation (u-8)(u+1) = 0 u=8 or u=-1 now that we have the equation for u, we can then solve for x if u = x^3=8 or u=x^3=-1 then x=8^(1/3) = 2 x=-1^(1/3) = -1 {2, -1} confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There were no discrepancies ------------------------------------------------ Self-critique Rating:3 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^2 - 3 x - sqrt(x^2 - 3x) = 2 let u = sqrt(x^2-3x) and u^2 = x^2-3x u^2 – u = 2 zero factor by adding -2 to both sides u^2 – u -2 =0 factor (u-2)(u+1) = 0 u =2 or u =-1 from here solve for x we know the square of a real number can not be a negative number if sqrt of x^2-3x =2 then x^2 -3x =4 add -4 to both sides x^2 -3x -4 =0 (x-4)(x+1) =0 x =4 x =-1 {-1, 4} confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 were no discrepancies ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^4 + sqrt(2) x^2 - 2 = 0 let u = x^2 and u^2 = x^4 u^2 + sqrt(2) I got lost here. I couldn’t figure out if the next step was to multiply u-2? the way it is in the original sqrt(2) x^2 doesn’t make sense to me. confidence rating: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I am still confused. I will take this one to my tutor also.