course mth 158 ǥGظm䇎n~{assignment #012
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02:06:25 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 (1-2x)^(1/3) = 1 1-2x=1 -2x=0 x=0
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02:06:36 ** 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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Ιڱcvˏ assignment #012 012. `query 12 College Algebra 07-10-2008
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13:20:43 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 (1-2x) = 1 2x+1=1 2x=0 x=0
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13:20:47 ** 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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13:24:49 **** 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. sqrt(3x+7) = 1.-sqrt(x+2) 3x+7= 1 - 2sqrt(x+2) +x+2 2x+4= - 2sqrt(x+2) 4x^2 + 16x + 16= 4x +4 4x^2+12x+12 4(x^2+3x+3) 4(x+2)(x+1) x= -2 or x= -1
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13:25:11 ** 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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13:28:53 **** 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. x^(3/4) = 9 x^(1/4) x^3=9x x(x^2)=x(9) x^2=9 x=3
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13:31:12 ** 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 --> ok i see x^(3/4) - 9 x^(1/4) = 0 x^(1/4) ( x^(1/2) - 9) = 0 x^(1/4) = 0 or x^(1/2) - 9 = 0 x^(1/2) = 9 x = 81 or x=0
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20:06:33 **** 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 x^3=a ax^3 - 7a -8 =0 ax^3-7a=8
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20:08:58 ** 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 is see my mistake a=x^3 so x^6 - 7x^3 - 8=0 a^2 - 7 a - 8 = 0 (a-8)(a+1) = 0 a = 8 or a = -1. x^3=8 or x^3= -1 x=2 or x= -1
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20:21:45 **** 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 --> x^2 - 3 x - sqrt(x^2 - 3x) = 2. x^4- 3 x^3 -(x^2 - 3x) = 4 x^2((x^2 - 3x) - (x^2 - 3x)=4 (x^2 - 3x)=a ax^2-a=4
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20:26:23 ** 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 i see what i did i needed to take x^2-3 out for a and i only took x^2 and i also should have left it sqrt u = sqrt(x^2 - 3x). u^2 - u = 2. u^2 - u - 2 = 0 (u-2)(u+1) = 0. u = 2 or u = -1.
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20:58:49 **** query 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.
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RESPONSE --> x^2=a a^2+ sqrt(2)a - 2=0 a^4+ 2a^2 -4=0 (a^2+2a-2)^2=0 a^2+2a-2=0
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21:03:26 ** 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 --> ok i really dont think I get this problem. Where did all the sqrts come from??
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21:03:34 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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