唔識做功課31

1)
2x² - px + q = 0has two real roots α and β. If α : β = 2 : 1, which of the following must be true?
1. p > 0
2. q > 0
3. 8q < p²

Solution :
α : β = 2 : 1
α = 2β

Sum of the roots:
α + β = p/2
2β + β = p/2
p = 6β

Product of the roots:
αβ = q/2
(2β)(β) = q/2
q = 4β²

1. p > 0 is false
p = 6β
when β < 0, p < 0

2. q > 0 is true
β is a non-zero real number, then β²> 0
Hence, q = 4β² > 0

3. 8q - p² < 0 is true
8q - p²
= 8(4β²) - (6β)²
= 32β² - 36β²
= -4β² < 0
Hence, 8q < p²


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2)
The roots of the quadratic equation x²+ ax + b = 0 are α and β, and the roots of the quadraticequation x² + bx + a = 0are α + 1and β + 1.Find the value of a + b.

The roots of x²+ ax + b = 0are α and β.
Sum of roots : α + β = -a ...... [1]
Product of roots: αβ = b ...... [2]

The roots of x² + bx + a = 0 are α + 1 and β + 1.
Sum of roots : (α + 1) + (β + 1) = -b ...... [3]
Product of roots : (α + 1)(β + 1) = a ...... [4]

From [3] :
(α + β) + 2 = -b ...... [5]

[5] - [1] :
a - b = 2 ...... [6]

[4] :
(α + 1)(β + 1) = a
αβ + (α + β) + 1 = a ...... [7]

[7] - [1] - [2] :
a - (-a) - b = 1
a + a - b = 1
2a - b = 1 ...... [8]

[8] - [7] :
a = -1

代入 [6] 中：
(-1) - b = 2
b = -3

a + b
= (-1) + (-3)
= -4

2014-10-03 23:23:23 補充：
In the above solution of Q.2 :

Combining 2 and 4 would give : "b = a + b + 1" but NOT "a = a + b + 1"

Therefore, a = -1 and b = -3 instead.

我打錯,本新係b我打左a ~.~