急~form4 MATH question

1.
4x² + 3(3h - 1)x = 27h
4x² + 3(3h - 1)x - 27h = 0
Let α and -α be the two roots of the given equation.

Sum of the roots: α + (-α) = -3(3h - 1)/4 ...... [1]
Product of the roots: α(-α) = -27h/4 ...... [2]

From [1]:
3h - 1 = 0
h = 1/3

Put h = 1/3 into [2]:
-α² = -27(1/3)/4
α² = 9/4
α = 3/2 ..or.. α = -3/2

Hence, the two roots are 3/2 and -3/2.


2.
(a)
x² + kx - 2 = 0
a and (a + 3) are the roots of the equation.

Sum of the roots: a + (a + 3) = -k ...... [1]
Product of the roots: a(a + 3) = -2 ...... [2]

From [2]:
a² + 3a + 2 = 0
(a + 1)(a + 2) = 0
a = -1 ..or.. a = -2

(b)
When a = -1:
-1 and 2 are the roots of the equation.
x = -1 ..or.. x = 2
x + 1 = 0 ..or.. x - 2 = 0
(x + 1)(x - 2) = 0
x² - x - 2 = 0
k = -1

When a = -2:
-2 and 1 are the roots of the equation.
x = -2 ..or.. x = 1
x + 2 = 0 ..or.. x - 1 = 0
(x + 2)(x - 1) = 0
x² + x - 2 = 0
k = 1

Hence, k = -1 ..or.. k = 1

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1) If a root of the equation 4x^2+3(3h-1)x=27h is the negative value of the other root and h is a constant,find the two roots.

Let α and β be roots of the equation

α + β = 0 = -3(3h-1)/4 ==> h = 1

αβ = -27h/4 = -27/4

since β = -α

α(-α) = -27/4

α = (1/2)√(27)

β = (-1/2)√(27)


2) If a and a+3 are the roots of the equation x^2+kx-2=0 and k is a constant, find the values of each of the following.
(a) a
(b) k

sum of roots a + (a+3) = -k

2a + 3 = -k

product of roots

(a)(a+3) = -2

2a + 3 = -k ; 2a = -k-3 = -(k+3)

a = -(k+3)/2

sub into (a)(a+3) = -2

a^2 + 3a + 2 = 0

[-(k+3)/2]^2 + 3[-(k+3)/2] + 2 = 0

[(k+3)^2]/4 -(3/2)(k+3) + 2 = 0

(k^2 + 6k + 9)/4 -(3/2)(k+3) + 2 = 0

全式乘 4

(k^2 + 6k + 9) - 6(k+3) + 8 = 0

(k^2 + 6k + 9) - 6k - 18 + 8 = 0

k^2 -1 = 0

k = ±1

a = -1 or a=-2

2013-10-12 16:14:17 補充：
sorry α + β = 0 = -3(3h-1)/4 ==> h = 1

計錯數，睇樓下啦！