course mth 158 7/9 1pm 012. `* 12
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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):OK ------------------------------------------------ Self-critique rating #$&*OK ********************************************* Question: * 1.4.28 (was 1.4.18). Explain how you found the real solutions of the equation sqrt(3x+7) + sqrt(x+2) = 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: sqrt (3x + 7) + sqrt (x+2) = 1 Make one radical per side sqrt (3x + 7) = -sqrt(x+2) +1 Square both sides sqrt (3x + 7)^2 = (-sqrt(x+2) + 1) ^2 3x + 7 = -sqrt(x+2)^2 - 2(sqrt(x+2)(1)) + 1 Simplify 3x + 7 = x+3 - sqrt2(x+2) subtract x + 3 from both sides. 2x + 4 = -sqrt2(x+2) Square both sides. (2x + 4)^2 = -sqrt2(x+2)^2 4x^2 + 16x + 16 = 4x + 8 subtract 4x + 8 from both sides. 4x^2 +12x + 8 = 0 Factor 4 (x^2 + 3x + 2) = 0 4 (x+1) (x+2) = 0 Divide 4 from both sidse (x+1) (x+2) = 0 x = -1 x = -2 x={-2} confidence rating #$&*2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique rating #$&*OK ********************************************* Question: * 1.4.40 (was 1.4.30). Explain how you found the real solutions of the equation x^(3/4) - 9 x^(1/4) = 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^(3/4) - 9x^(1/4) = 0 Factor out x^(1/4) x^(1/4) (x^1/2 - 9) = 0 Zero Property x^(1/4) = 0 (x^1/2-9) = 0 x^(1/2) = 9 Square x^(1/2)^2 = 9 ^2 x = 81 x= {0, 81} confidence rating #$&*2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique rating #$&*OK ********************************************* Question: * 1.4.46 (was 1.4.36). Explain how you found the real solutions of the equation x^6 - 7 x^3 - 8 =0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Reworked: Substitute a value for x. x^6 - 7x^3 - 8 = 0 a^2 - 7a - 8 = 0 Factor a^2 - 8a + 1a - 8 = 0 a(a - 8) + (a - 8) = 0 (a - 8) (a + 1) = 0 Zero Property a = 8, a = -1 substitute original x. x^3 = 8, x^3 = -1 x = 2, x = -1 {2, -1} confidence rating #$&*0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I honestly had no idea how to approach this problem. Reworking after I realized you could substitute a for x^3, I had no problems. ------------------------------------------------ Self-critique rating #$&*2 ********************************************* Question: * 1.4.64 (was 1.4.54). Explain how you found the real solutions of the equation x^2 - 3 x - sqrt(x^2 - 3x) = 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You can substitute a = sqrt x^2 - 3x a^2 - a = 2 a^2 - a -2 =0 Factor to (a + 1) (a-2) a = -1, a = 2 Substitute for a sqrt (x^2 - 3x) = -1 is impossible. Square both sides sqrt (x^2 -3x) = 2 x^2 - 3x = 4 Factor to (x +1) (x - 4) =0 x = -1, x = 4 {-1, 4} confidence rating #$&*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): OK ------------------------------------------------ Self-critique rating #$&*OK ********************************************* Question: * 1.4.92 \ 90 (was 1.4.66). Explain how you found the real solutions of the equation x^4 + sqrt(2) x^2 - 2 = 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Subsitute a = x^2 a^2 + sqrt (2) a - 2 = 0 Solve using the quadratic formula. a = [-b +- sqrt (b^2 - 4ac)] / 2a a = [-sqrt2 +- sqrt (2^2 - 4(1)(-2)]/2(1) a = [-sqrt2 +- sqrt (2 - (-8)) / 2 a = (-sqrt2 +- sqrt10) / 2 Substitute x^2 for a x^2 = (-sqrt2 + sqrt 10) / 2. x = sqrt (-sqrt(2) + sqrt(10))/2) x = -sqrt ( -sqrt(2) + sqrt(10))/2) x = .935, x = -.935 confidence rating #$&*2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique rating #$&*OK Add comments on any surprises or insights you experienced as a result of this assignment."