Prelim 2016 Maths Help Thread (1 Viewer)

[Green Yoda]

NOTE I AM WRITING a as alpha and b as beta
(a+b)=3, (ab)=-13
b)
cross multiply 1/a^2 + 1/b^2
to get (a^2+b^2)/(a^2*b^2)
=((a+b)^2-2ab))/(ab)(ab)
=((3)^2-2(-13))/(-13)(-13)
=35/169

 

[jathu123]

Parts 2 and 3 are what I'm stumped on.

Part 3: If you sub b into the original equation, it must turn out to be 0 (since b is a root). So b^2-3b-13=0. Now multiply each side by -2, you get -2b^2+6b+26=0. Now subtract 26 from both sides and rearrange the LHS to get 6b-2b^2=-26. So you can see the LHS is exactly what the question wants you to find, so the answer is just -26

yeh I'm just using alpha and beta as a and b, ceebs

 

[Kreisler]

Help pls

 

[InteGrand]

Help pls

 

[Orwell]

 

[InteGrand]

 

[Orwell]

From which God were you birthed Integrand?

 

[Kreisler]

Help again pls ;_;

 

[Kreisler]

Sorry sent same question in twice : /

 

[Orwell]

Hippity hoppity helpitus.

 

[Paradoxica]

Help again pls ;_;

Let the roots be a, 2a

Then the sum of the roots: 3a = -m

Product: 2a² = k

9a² = m²

18a² = 2m²

9k = 2m²

------------------------------

x² - px + p = 0

The roots are opposite in sign if and only if the product of the roots is a negative number.

The product of the roots is p, but the question stated that p is positive. Therefore, the roots are never opposite in sign.

=============================

Δ = p² - 4p

The roots exist if and only if Δ ≥ 0

p² - 4p ≥ 0

p² -4p + 4 ≥ 4

(p-2)² ≥ 2²

p-2 ≥ 2 OR p-2 ≤ -2

p ≥ 4 OR p ≤ 0

p ≥ 4 since p > 0

 

[Paradoxica]

Hippity hoppity helpitus.

Same approach as Integrand...

4(1-sin²θ) = 6sinθ + 6

6(1+sinθ) - 4(1-sinθ)(1+sinθ)= 0

(2-4sinθ)(1+sinθ) = 0

You can carry the rest...

------------------------------------

(x+1)² = 8y-24 = 8(y-3)

 

[Kreisler]

Help me pls

 

[Orwell]

 

[Green Yoda]

Q80
i)
(a+b)=4
(ab)=2

ii)
3(a+b)=12=-b/a
3a*3b=9ab=18=c/a
a=0 b=-12 c=6
new eqn = x^2-12x+6

81)
if (a+b)=0 then -b/a =0 therefore -b=0
therefore -(1+m)=0
m=1

 

[trecex1]

Q80i) just use -b/a and c/a
ii) Make a monic quadratic in form x^2 -(3α+3β) +9αβ (use part i))
Q81, use sum of roots equate them to 0.
These are pretty intuitive quadratic function questions to be honest.

 

[Green Yoda]

Help me pls

Q77
i) complete the square
x^2-2x=-4y-9
(x-1)^2=-4y-8
(x-1)^2=-4(y+2)

ii) It is an vertical negative parabola vertex at (1,-2), focal length 1 so the focus is at -1 units from vertex (1,-3) by inspection
iii) directrix is +1 units from vertex, so its y=-1

Q78
i)
Δ=4k^2-4(8k-15)
4k^2-32k+60
k^2-8k+15

ii)
Δ=k^2-8k+15>=0
(k-3)(k-5)>=0
k<=3 or k>=5

iii)
3(2k)=2(8k-15)
6k=16k-30
10k=30
k=3

 

[Green Yoda]

What is the answer to Q80(ii)? I am unsure about this one.

 

[InteGrand]

What is the answer to Q80(ii)? I am unsure about this one.

 

[Green Yoda]

would u then just sub b as (3a+3b) and c as 9ab?
to get x^2-(3a+3b)+9ab? or do we need numerical results?