Assignment 12

This looks good. See my notes on a couple of errors, but your approach on almost all problems was correct.

Let me know if anything is unclear, and include specifics about what you do and do not understand.

......!!!!!!!!...................................

20:45:52 query 1.4.12 (was 1.4.6). Explain how you found the real solutions of the equation (1-2x)^(1/3) - 1 = 0

......!!!!!!!!...................................

RESPONSE --> cube rt (1-2x) - 1 = 0 isoloate radical by adding one to each side cube rt (1-2x) = 1 raise each side to third power cube rt (1-2x)^(3) = 1^3 -2x + 1 = 1

so far so good, but the next step is incorrect

To get this you would have to add 1 to the left side and subtract 1 from the right side.

.................................................

......!!!!!!!!...................................

20:49:25 ** 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. **

......!!!!!!!!...................................

RESPONSE --> I see. I tried to subtract 1 to get the equation equal to zero and solve rather than simply doing it this way.

.................................................

......!!!!!!!!...................................

20:56:42 **** 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.

......!!!!!!!!...................................

RESPONSE --> sqrt (3x+7) = 1 - sqrt(x+2) sqrt (3x+7)^(2) = (1 - sqrt (x+2))^(2) 3x+7 = x +2 - 2 sqrt (x +2) + 1 2x + 4 = -2sqrt(x + 2) (2x + 4)^2 = (-2sqrt(x+2))^(2) 4x^2 + 16x +16 = 4(x+2) 4x^2 + 16x + 16 = 4x + 8 4x^2 + 12x + 8 = 0 4(x^2 + 3x + 2)=0 4(x+2)(x+1)=0 x=-2,-1 solution is -2 because -1 doesn't work when checked

.................................................

......!!!!!!!!...................................

20:57:28 ** 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. **

......!!!!!!!!...................................

RESPONSE --> right

.................................................

......!!!!!!!!...................................

21:01:57 **** 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.

......!!!!!!!!...................................

RESPONSE --> x^(3/4) = 9x^(1/4) [x^(3/4)]^(4) = [9x^(1/4)]^(4) x^3 = 9x

This doesn't follow because you would also have to raise 9 to the 4th power. Your equation would be x^3 = 9^4 * x.

.................................................

......!!!!!!!!...................................

21:05:19 ** 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). **

......!!!!!!!!...................................

RESPONSE --> I did this one completely differently. I am confused: if my solutions check as well as yours, am I doing this correctly?

See my note.

Had you plugged 3 into the original equation it would not have checked out. Also, note that the 1/4 power of -3 would not be a real number.

.................................................

......!!!!!!!!...................................

21:09:56 **** 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

......!!!!!!!!...................................

RESPONSE --> let u = x^3 change the expression substituting with u and solve. then solve for x with values of u u^2 - 7u - 8 = 0 (u-8)(u+1)=0 u=8, -1 x^3 = 8 cube rt(x^3) = cube rt (8) x=2 x^3 = -1 cbrt (x^3) = cube root (-1) x=-1 When checked, 2 is the only real solution.

.................................................

......!!!!!!!!...................................

21:10:18 ** 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. **

......!!!!!!!!...................................

RESPONSE --> right

.................................................

......!!!!!!!!...................................

21:18:48 **** 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.

......!!!!!!!!...................................

RESPONSE --> let u = sqrt (x^2 - 3x) u^2 - u = 2 u^2 - u - 2 = 0 (u-2)(u+1)=0 u = 2,-1 for solution u = 2 sqrt (x^2 - 3x) = 2 [sqrt (x^2 -3x)]^2 = 2^2 x^2 - 3x =4 x^2 - 3x -4 =0 (x-4)(x+1)=0 x = 4,-1 for solution u = -1 sqrt (x^2 - 3x) = -1 [sqrt(x^2 - 3x)]^2 = (-1)^2 x^2 - 3x +1 = 0 prime, does not factor after checking all solutions, the only real solution is 4

.................................................

......!!!!!!!!...................................

21:20:00 **** 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.

......!!!!!!!!...................................

RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. ok

.................................................

......!!!!!!!!...................................

21:31:25 ** 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. **

......!!!!!!!!...................................

RESPONSE --> I apologize. I must have entered twice without seeing a new problem entered. I see this is 90 from our assignment. I am glad you chose this one because I am unsure about something. Where did u=(-sqrt(2) + sqrt (10) )/2 come from? When working it out, it only comes up to a negative sqrt of 10.

The quadratic formula tells you that u = ( - b +- sqrt(b^2 - 4 a c) ) / (2 a). The + sqrt(10) comes from the quadratic formula, as does the -sqrt(10).

.................................................

......!!!!!!!!...................................

21:31:57 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.

......!!!!!!!!...................................

RESPONSE --> This wasn't very complicated for me.

.................................................