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June 8th, 2014, 12:02 AM   #1
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Post Maths Assignment help!!?

Please help me to complete my maths assignment

Signs-
√ = square root
² = square
x¯ = bar ( 0.7¯ = 0.77777777...) (0.47¯¯=0.47474747..)
^ = raised to power of eg. 4^-2 = 1/16



*************
1. if a = 8+3√7 and b =1/a then find a² +b²

2. Express 0.6 + 0.7¯ + 0.47¯¯ in the form p/q where p and q are integers and q is not equal to 0

3.Rationalise 1 / √ 3 +√ 2 -√ 5

4. Find the value of a and b in √ 7-1 / √ 7+1 - √ 7+1 / √ 7-1 = a +b√ 7

5. If x, y and z are real numbers show that √ x^-1 √y * √y^-1 √z * √z^-1 √x =1

6. (i) a+b =11 , ab =28 find the value of a³ + b³
(ii) if x² + 1/x² =18 find the value of x³ - 1/x³

7. (i) if a+b+c = 5 and ab+bc+ca= 10 , prove that a³ + b³ +c³-3abc = -25
(ii) If a , b , c are all non-zero and a+b +c =0 then prove that a²/bc + b²/ca + c²/ab =3

8. Find the value of a and b so the polynomial x³ + 10x² + ax + b is exactly divisible by (x-1) 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.
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June 8th, 2014, 03:17 AM   #2
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1. 1/(8 + √7) = (8 - √7)/((8 - √7)(8 + √7)) = (8 - √7)/57.
Also, (8 ± √7)² = 71 ± 16√7, etc.

2. 3/5 + 7/9 + 47/99 = 917/495

3. 1/(√3 + √2 - √5) = (√3 + √2 + √5)/(3 + 2√6 + 2 - 5) = (3√2 + 2√3 + √30)/12

4. (√7 - 1)/(√7 + 1) - (√7 + 1)/(√7 - 1) = (√7 - 1)²/6 - (√7 + 1)²/6 = -2√7/3

5. Can't be done, as the result is false for x = y = z = -1.

6.(i) a³ + b³ = (a + b)³ - 3ab(a + b) = 11³ - 3(28)(11) = 407
6.(ii) By (i), x^6 + 1/x^6 = (x² + 1/x²)³ - 3(x² + 1/x²) = 5778,
so (x³ - 1/x³)² = x^6 + 1/x^6 - 2 = 5776 = (±76)².

7.(i) a³ + b³ + c³ - 3abc = (a + b + c)³ - 3(a + b + c)(ab + bc + ca) = -25
Thanks from firasjavid

Last edited by skipjack; June 8th, 2014 at 03:39 AM.
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June 8th, 2014, 03:23 AM   #3
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Thanks but what about Q8?

...
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June 8th, 2014, 06:52 AM   #4
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Hello, firasjavid!

Quote:
8. Find the value of $a$ and $b$ so the polynomial $x^3 + 10x^2 + ax + b$
is exactly divisible by $(x-1)$ as well as by $(x+2).$

Let $p$ be the third root of the cubic.

Then: $\:(x-p)(x-1)(x+2) \:=\:x^3 + 10x + ax + b$

$\qquad x^3 + (p+1)x^2 + (p-2)x - 2p \:=\:x^3 + 10x + ax + b$

Equate coefficients: $\:\begin{Bmatrix}p+1 &=& 10 \\ p-2 &=& a \\ -2p &=& b \end{Bmatrix}$

Therefore: $\;p \:=\: 9 \quad\Rightarrow\quad \begin{Bmatrix}a &=& 7 \\ b &=& \text{-}18 \end{Bmatrix}$

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June 8th, 2014, 10:23 AM   #5
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Math Focus: Elementary mathematics and beyond
f(x) = x³ + 10x² + ax + b

f(1) = 0 and f(-2) = 0, hence -2a + b = -32 and a + b = -11, a = 7, b = -18.
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June 8th, 2014, 12:28 PM   #6
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Quote:
Originally Posted by firasjavid View Post
Please >>>> help me to complete <<< my maths assignment
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 take-home
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.
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June 9th, 2014, 06:10 AM   #7
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Quote:
Originally Posted by Math Message Board tutor View Post
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 take-home
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.
Welcome, lookagain
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