course Mth 158 Ü^ë®ÔË¢ÒËèΫÆÎðMÔ•~Üúöª|Îassignment #012 012. `query 12 College Algebra 06-28-2008
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08:11:39 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 --> Add 1 to each side (1-2x)^1/3 = 1 Raise to the 3rd power leaving 1-2x =1 Add -1 to each side -2x=0 x=0
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08:11:49 ** 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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08:45:18 **** 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 --> Isolate the largest variable by substracting (sqrt x +2) from each side leaving sqrt of 3x+7= 1- (sqrt x +2) Square each side sqrt (3x+7)^ 2 = (-sqrt (x+2) + 1 ^2 3x+7= 1+ 2sqrt x +2 +x+2 3x +7 =x+3 -2 sqrt (x+2) 3x+7-X-3 = -2sqrt (x+2) 2x+4 = -2sqrt (x+2) Square each side 4x^2 +16x +16 = 4(x+2)Add -4x +8 to both sides 4x ^2 ++8+12x =0 Factor 4 (x+1) (x+2) =0 (x+1) (x+2) = 0 x+1 = 0 or x+2=0 x= -1 or x=-2
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08:45:48 ** 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 need to check the solutions to see which ones actually work.
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08:51:21 **** 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 --> Factor to get (x ^ 1/4) (x^ (1/2) -9 = 0 x^ 1/4 = 0 or x^ 1/2 -9 = 0 x= 0 or x=81
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08:51:26 ** 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
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08:58:12 **** 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) x^3 -8 = 0 or x^3 +1 = 0 x^3 = 8 or x^3 = -1 x=2 or x=-1
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08:58:52 ** 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 --> Do you have to change the equation or can you just factor like I did
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09:08:03 **** 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 u = sqrt x^2 - 3x Let u^2 = x^x-3x giving us the equation u^ 2 -u -2 =0 Factor (u-2) (u+2) = 0 u= 2 or u= -1 Put back into equation sqrt (x^2-3x) =2 and sqrt (x^2 -3x) = -1 sqrt (x^2 -3x) =2 Substract 2 so the equation will equal 0 X^2 -3x -2 =0 Factor (x-4) (x+1) = 0 x= 4 or x= -1
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09:08:34 ** 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 --> Why is the answere -4 and 1 and not 4 and -1
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10:12:57 **** 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 --> Let u = X^2 so that u2 =x^4 u^2 = sqrt 2u -2=0 u=-sqrt2+-sqrt of the sqrt 3^2- 4(1)=-sqrt I am lost on this one
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10:17:44 ** 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 use the wuadratic formula to get the equation u= (-sqrt 2 +- sqrt (10)/2 Simplify u= (-sqrt 2 +- sqrt (10)/2 x^2 = (-sqrt (2) = sqrt (10)/2 and then finish out with a calculator. Ok
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