F.4 Maths 20POINTS!!!!!!!!

1.
The answer is : 1, 2

1. true
Discriminant Δ = 1² - 4*(-1)*(-4) = -15 < 0
Hence, the equation has no real roots.

2. true
Discriminant Δ = 1² - 4*(-1)*(-2) = -7 < 0
Hence, the equation has no real roots.

3. false
Discriminant Δ = 1² - 4*(-1)*1 = 5 > 0
Hence, the equation has two distinct real roots.


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2.
The answer is : a) ac < 0

a) true
Disriminant Δ = b² - 4ac
b² must be ≥ 0.
When ac < 0 : -4ac > 0, and thus b² - 4ac > 0
Hence, the equation has two unequal roots.

b) false
Disriminant Δ = b² - 4ac
b² must be ≥ 0.
When ac > 0 : -4ac < 0
Then, Δ ≥ 0 when b² ≥ -4ac. Otherwise, Δ < 0
Hence, the nature of the roots cannot be determined.

c) false
When Δ = b² - 4ac < 0, the equationhas no real roots.

d) false
When Δ = b² - 4ac = 0, the equation hastwo equal roots.


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3.
(x + 3)(x + 4) = x + p
x² + 7x + 12 = x + p
x² + 6x + (12 - p) = 0

The equation has equal roots. Then, discriminant Δ = 0
6² - 4*1*(12 - p) = 0
36 - 48 + 4p = 0
4p = 12
p = 3


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4.
(2x² - 4x + 5) + k(x - 1) = 0
2x² - 4x + 5 + kx - k = 0
2x² + (k - 4)x + (5 - k) = 0

The equation has two equal roots. Then, discriminant Δ = 0
(k - 4)² - 4*2*(5 - k) = 0
k² - 8k + 16 - 40 + 8k = 0
k² = 24
k = √24 or k = -√24
k = 2√6 or k = -2√6


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5.
α and β are the roots of x² + 3x − 4 = 0
Sum of the roots : α + β = -3

2^α•2^β
= 2^(α + β)
= 2^-3
= 1/2^3
= 1/8


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6.
α and β are the roots of ax² + bx + c = 0
Sum of the roots : α + β = -b/a
Product of the roots : αβ = c/a

(α + 1/β) (β + 1/α)
= [(αβ + 1)/β] [(αβ + 1)/α]
= (αβ + 1)²/αβ
= [(c/a) + 1]² / (c/a)
= [(a + c)²/a²]/ (ac/a²)
= (a + c)² / ac


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7.
Let 2k and 3k be the roots of the equation 3x² + bx + 32 = 0

Product of the roots :
2k * 3k = 32/3
6k² = 32/3
k² = 16/9
k = 4/3 or k = -4/3

Sum of the roots :
2k + 3k = -b/3
b = -15k
b = -15*(4/3) or b = -15*(-4/3)
b = -20 or b = 20

1. For 3,
1 + x - x^2 = 0
x^2 - x - 1 = 0
(x - 1)(2x + 1) = 0
x = 1 or -1/2


2. c and d must be wrong, cuz b^2-4ac must be >0 if there are 2 unequal real roots.
ac>0 may be correct, but not in all cases.
Therefore the answer is A.


3. (x + 3)(x + 4) = x + p
x^2 + 7x + 12 = x + p
x^2 + 6x + (12 - p) = 0

b^2-4ac = 0 (equal real roots)
6^2 - 4(1)(12 - p) = 0
36 - 48 + 4p = 0
4p =12
p = 3


4. (2x^2 - 4x + 5) + k(x −1) = 0
2x^2 - 4x + 5 + kx - k = 0
2x^2 + x(k - 4) + (5 - k) = 0

b^2-4ac = 0 (equal real roots)
(k - 4)^2 - 4(2)(5 - k) = 0
k^2 - 8k + 16 - 40 + 8k = 0
k^2 - 24 = 0
k = +2sqrt(6) or -2sqrt(6)


5. x^2 +3x − 4 = 0
(x - 1) (x + 4) = 0
x = 1 or -4

2^α ⋅ 2^β
= 2^1 ⋅ 2^(-4)
= 1/8


6. The calculation is quite similar to #5, so plz try yourself!! :)


7. 3x^2 + bx + 32 = 0

Let α and β be the roots.
Consider Ax^2 + Bx + C = 0
Sum of roots = -B/A
Product of roots = C/A
α + β = -b/3 -----------(1)
α ⋅ β = 32/3 -----------(2)

By ratio,
3α = 2β
α = 2β/3 --------------(3)

Sub. (3) into (1) and (2),
2β/3 + β = -b/3 -------------(4)
2β/3 ⋅ β = 32/3 -------------(5)

By solving the 2 equations (4) and (5)
(simultaneuos linear equations in 2 unknowns),
you can get the answer.

2012-10-07 17:31:27 補充：
Btw, thanks for helping me to do revision on my maths too:D
Add oil lah~