i cant find the 3rd root.. i have da marking scheme but doesnt explain it properly.. how do u get da 1/2? can some1 pleeease explain?!!!
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P4 Jan 05 QS 2a watch
- Thread Starter
- 25-06-2005 23:36
- 25-06-2005 23:52
well the other root is 3 - i
so (x - (3-i))(x - (3+i)) = x² - 6x + 10 is a factor of 2x³ + ax² + bx - 10
2x³ + ax² + bx - 10 = (nx + m) ( x² - 6x + 10 )
so 2x³ + ax² + bx - 10 = nx³ + ( m - 6n) x² + (10n-6m)x + 10m
so 10m = -10 , m = -1
2x³ = nx³
so n = 2
so 2x³ + ax² + bx - 10 = ( 2x - 1 ) ( x² - 6x + 10 )
so the third root is 1/2
- 26-06-2005 00:02
Alternatively, suppose the three roots are a, b and c. Then:
k(x-a)(x-b)(x-c) 2x^3 + ax^2 + bx - 10
After you multiply this out, you're left with 2 conclusions:
(1) k = 2
(2) -kabc = -10
abc = 10/k = 5
That is, the product of the 3 roots is 5. So:
(3+i)(3-i)c = 5
(9 - 3i + 3i + 1)c = 5
c = 5/10 = 1/2
And a third method would be to find a and b (from the next part), and use them to factorize the equation.
- 26-06-2005 06:28
ANYONE know how to do 4bii)?
i've drawn a tangent at 5f(6), but dont really know what to expect for x2 and x3....if there is no root in the interval does the gradient increase, always? Also, if x1 happened to be a turning point, which way would the second point go? or would that depend on the walue of f(x)/f'(x), and no way of predicting it
and for qeutsion 1, i got two answers...x<-3a and x<a/3, i didn't think about stating that x<a/3 satisfies both inequalities, so in exam situation would i get full marks for that solution?
- 26-06-2005 09:40
see attachment for 4bii,
for question 1, you lose a mark for not stating that x<a/3 is the final answer.
by the way, i think you have a typo: "x<-3a and x<a/3, i didn't think about stating that x<a/3 satisfies both inequalities," x<-3a satisfies both inquality not x<a/3