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June 8th, 2014, 12:02 AM  #1 
Newbie Joined: Jun 2014 From: Kuwait Posts: 8 Thanks: 0  Maths Assignment help!!?
Please help me to complete my maths assignment Signs âˆš = square root Â² = square xÂ¯ = bar ( 0.7Â¯ = 0.77777777...) (0.47Â¯Â¯=0.47474747..) ^ = raised to power of eg. 4^2 = 1/16 ************* 1. if a = 8+3âˆš7 and b =1/a then find aÂ² +bÂ² 2. Express 0.6 + 0.7Â¯ + 0.47Â¯Â¯ in the form p/q where p and q are integers and q is not equal to 0 3.Rationalise 1 / âˆš 3 +âˆš 2 âˆš 5 4. Find the value of a and b in âˆš 71 / âˆš 7+1  âˆš 7+1 / âˆš 71 = a +bâˆš 7 5. If x, y and z are real numbers show that âˆš x^1 âˆšy * âˆšy^1 âˆšz * âˆšz^1 âˆšx =1 6. (i) a+b =11 , ab =28 find the value of aÂ³ + bÂ³ (ii) if xÂ² + 1/xÂ² =18 find the value of xÂ³  1/xÂ³ 7. (i) if a+b+c = 5 and ab+bc+ca= 10 , prove that aÂ³ + bÂ³ +cÂ³3abc = 25 (ii) If a , b , c are all nonzero and a+b +c =0 then prove that aÂ²/bc + bÂ²/ca + cÂ²/ab =3 8. Find the value of a and b so the polynomial xÂ³ + 10xÂ² + ax + b is exactly divisible by (x1) as well as by (x+2) *************** Thanks if you helped. I really needed this. If some are confusing, just skip them. Last edited by skipjack; June 8th, 2014 at 03:15 AM. 
June 8th, 2014, 03:17 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,809 Thanks: 2150 
1. 1/(8 + âˆš7) = (8  âˆš7)/((8  âˆš7)(8 + âˆš7)) = (8  âˆš7)/57. Also, (8 Â± âˆš7)Â² = 71 Â± 16âˆš7, etc. 2. 3/5 + 7/9 + 47/99 = 917/495 3. 1/(âˆš3 + âˆš2  âˆš5) = (âˆš3 + âˆš2 + âˆš5)/(3 + 2âˆš6 + 2  5) = (3âˆš2 + 2âˆš3 + âˆš30)/12 4. (âˆš7  1)/(âˆš7 + 1)  (âˆš7 + 1)/(âˆš7  1) = (âˆš7  1)Â²/6  (âˆš7 + 1)Â²/6 = 2âˆš7/3 5. Can't be done, as the result is false for x = y = z = 1. 6.(i) aÂ³ + bÂ³ = (a + b)Â³  3ab(a + b) = 11Â³  3(28)(11) = 407 6.(ii) By (i), x^6 + 1/x^6 = (xÂ² + 1/xÂ²)Â³  3(xÂ² + 1/xÂ²) = 5778, so (xÂ³  1/xÂ³)Â² = x^6 + 1/x^6  2 = 5776 = (Â±76)Â². 7.(i) aÂ³ + bÂ³ + cÂ³  3abc = (a + b + c)Â³  3(a + b + c)(ab + bc + ca) = 25 Last edited by skipjack; June 8th, 2014 at 03:39 AM. 
June 8th, 2014, 03:23 AM  #3 
Newbie Joined: Jun 2014 From: Kuwait Posts: 8 Thanks: 0  Thanks but what about Q8?
...

June 8th, 2014, 06:52 AM  #4  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, firasjavid! Quote:
Let $p$ be the third root of the cubic. Then: $\:(xp)(x1)(x+2) \:=\:x^3 + 10x + ax + b$ $\qquad x^3 + (p+1)x^2 + (p2)x  2p \:=\:x^3 + 10x + ax + b$ Equate coefficients: $\:\begin{Bmatrix}p+1 &=& 10 \\ p2 &=& a \\ 2p &=& b \end{Bmatrix}$ Therefore: $\;p \:=\: 9 \quad\Rightarrow\quad \begin{Bmatrix}a &=& 7 \\ b &=& \text{}18 \end{Bmatrix}$  
June 8th, 2014, 10:23 AM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond 
f(x) = xÂ³ + 10xÂ² + ax + b f(1) = 0 and f(2) = 0, hence 2a + b = 32 and a + b = 11, a = 7, b = 18. 
June 8th, 2014, 12:28 PM  #6 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  None of you tutors/moderators should have posted any answers for this lazy user. He/she showed zero work, but used the phrase highlighted above. This lazy user also posted this same set of homework problems (or takehome quiz/test problems) on other math message board sites. Only when the user has made an attempt and has shown some work on a forum toward a certain problem should a worked out (or partially worked out) solution be given. This is an unofficial universal etiquette regardless of which math message board someone is on, but it is enforced on certain ones over others. 
June 9th, 2014, 06:10 AM  #7  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  Quote:
 

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