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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,344

A # 1

If x² + y² + z² = 50, xy + yz + xz = 47 and xyz = 60,

what is the value of

(i) x

(ii) y and

(iii) (x+y+z)³?

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

the numbers are {3,4,5}

12^3=1728

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,344

Character is who you are when no one is looking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Thanks.

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**ganesh****Moderator**- Registered: 2005-06-28
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A # 2

For waht value of m is 2x³-x²-3mx-24 divisible by x-2?

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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m=2?

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**ganesh****Moderator**- Registered: 2005-06-28
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|m|=2 would have been better! The answer is m=-2. Try again, krassi_holmz, probably you made a mistake somewhere.

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**ganesh****Moderator**- Registered: 2005-06-28
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A # 3

If ax³+bx²+x-6 has (x+2) as a factor and leaves a remainder 4 when divided by (x-2), find the values of a and b.

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**ganesh****Moderator**- Registered: 2005-06-28
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A # 4

The sum of two natural numbers is 8. Determine the numbers if the sum of their reciprocals is 8/15.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

A # 3

From the information given, we can deduce that f(-2) = 0 and f(2) = 4.

-8a + 4b -8 = 0

8a + 4b -4 = 4

Subtract the two equations:

16a -4 = -4

16a = 0

a = 0.

Add the two equations: 8b -12 = 4

8b = 16

b = 2.

**f(x) = 2x²+x-6**

A # 4

Calling the 2 numbers p and q, we know that p+q = 8, and that 1/p+1/q = 8/15.

Cross-multiplying the second equation gives (p+q)/pq = 8/15. We know that (p+q) = 8, so pq must be 15.

The numbers that add to give 8 and multiply to give 15 are 3 and 5.

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**ganesh****Moderator**- Registered: 2005-06-28
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Well done, mathsyperson!

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**ganesh****Moderator**- Registered: 2005-06-28
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A # 5

If one of the roots of the equation x²-3x+q=0 is twice the other, find the value of q.

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**ganesh****Moderator**- Registered: 2005-06-28
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A # 6

If the ratio of the roots of the equation ax²-2ax+b=0 is equal to that of ax²-2cx+d=0, then which of the following is true?

(a) a²b=c²d (b) a²c=b²d

(c) a²d=c²b (d) d²b=c²a

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**ganesh****Moderator**- Registered: 2005-06-28
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A # 7

Solve: (x²+1/x²) - 3(x-1/x)-2=0

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**amonjezi****Member**- Registered: 2006-03-09
- Posts: 9

y=4

and,y=-1

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**amonjezi****Member**- Registered: 2006-03-09
- Posts: 9

(x

+1/x^2)-3(x-1/x)-2+2-2=0

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**amonjezi****Member**- Registered: 2006-03-09
- Posts: 9

(x^2+1/x^2)-3(x-1/x)-2+2-2=0 then (x+1/x)^2-3(x-1/x)-4=0 if (x+1/x)=y y=4 and -1 now we solves (x+1/x)=4 and (x+1/x)=-1 these equaition very easy

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**amonjezi****Member**- Registered: 2006-03-09
- Posts: 9

hi ganesh

choice (a) or(c) are true

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**Eeyore****Member**- Registered: 2007-04-16
- Posts: 8

a7 1 and -1

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**Eeyore****Member**- Registered: 2007-04-16
- Posts: 8

A#5 q=2

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 91,413

Hi ganesh;

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**soroban****Member**- Registered: 2007-03-09
- Posts: 452

. .

.

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**soroban****Member**- Registered: 2007-03-09
- Posts: 452

.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,344

Hi soroban,

I too get the same solutions. Well done!

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**G-man****Member**- Registered: 2011-02-28
- Posts: 16

Maths!......

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