course Mth 158 r?????????????assignment #012
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20:41:48 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 --> (1-2x)^(1/3) - 1 = 0 add 1 to both sides.. (1-2x)^(1/3)-1+1=1 (1-2x)^(1/3)=1 raise each side to the 3rd power.. (1-2x)^(1/3)^3=1^3 1-2x=1 so that.. -2x=1-1.. -2x=0 (x)/(-2)=(0)/(-2) x=0
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20:42:08 ** 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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20:46:51 **** 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 --> sqrt(3x+7) + sqrt(x+2) = 1. raise each side to the 2nd power..sqrt(3x+7)^2 + sqrt(x+2)^2= 1^2 then you get.. 3x+7+x+2=1 add and subtract like terms..4x+9=1 subrtact 9 from both sides.. 4x=-8 divide both sides by 4..(4x)/(4)=(-8)/(4) sp that x=-2.. no real solution
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20:48:53 ** 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 did this problem the other way.. and still recieved the same answer
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21:02:05 **** 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 --> x^(3/4) - 9 x^(1/4) = 0 move -9x^(1/4) to the opposite side to balance the equation.. x^(3/4)=9x^(1/4) raise each side to the 4th power x^(3/4)^4=9x^(1/4)^4 x^3=9x move 9x over to the opposite side so you can factor.. x^3-9x=0 x(x^2-9)=0 factor again... x(x-3)(x+3)=0 so that x=0.. and.. x-3=0.. x=3 and.. x+3=0...x=-3 concluding.. x=0, x= -3, x=3
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21:04:57 ** 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 understand.. i should have factored (1/4) first
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21:11:41 **** 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 --> x^6 - 7 x^3 - 8 =0 factor.. (x^3-8)(x^3+1) apply zero property.. x^3-8=0 x^3+1=0 x^3=8.. and x^3=-1 take the 3rd root of each.. x^3= -8..x= -2 and.. x^3= -1.. x= -1 so that x= -2 and x=-1.. {-2,-1}
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21:13:57 ** 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 think the only thing I did wrong was getting -2 for the 3rd root of -8.. I did the problem differently but I think i basicaly did the same thing this solution did
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21:27:45 **** 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 --> let c= sqrt (x^2 - 3x) and c^2=x^2-3x c^2+c=2 subtract 2 from both sides.. c^2+c-2=0 factor.. (c+2)(c-2)=0 c= -2 and c= 2 c= -2..so that sqrt(x^2-3x)= -2.. (can't be done because a principal square root is always non-negative.. or c= 2.. so that sqrt(x^2-3x)=2 square both sides to get.. x^2-3x=4 subtract 4 from both sides.. x^2-3x-4=0 factor.. (x-4)(x+1)=0 x=4 or x= -1
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21:28:09 ** 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
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21:54:59 **** 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 --> x^4 + sqrt(2) x^2 - 2 = 0 let m=x^2..m^2=x^4 m^2+sqrt(2)m-2=0 m= -sqrt((2) + or - sqrt(2+8)) / (2) m= -sqrt(2) + or - sqrt(10) / 2 m= -sqrt(2) + sqrt(10) / 2 x^2= -sqrt(2) + sqrt(10) / 2 x= + or - sqrt of ((-sqrt(2) + sqrt(10)) / 2 or m= -sqrt(2) - sqrt(10) / 2.. so that x^2= -sqrt(2) - sqrt(10) / 2 x= + or - sqrt of ((-sqrt(2) - sqrt(10)) / 2 (not a real number) therefore, x= sqrt of ((-sqrt(2) + sqrt(10)) / 2
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21:56:12 ** 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 forgot to turn the answers into decinmals.. where I would have gotten two answers if done so.. but my answer checks out with this answer before done by a calculator
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21:56:13 07-03-2006 21:56:13 ** 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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NOTES ------->
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21:56:27 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> ok
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