20分!!F.4 QUADRATIC EQUATION 4條

網路上的路人甲

2010-12-29 6:48 am

1.It is given that a and b are the roots of the equation x^2+kx-2=0, where k>0.
if a^2+b^2=13,

a)find the value of k.
b)find a quadratic equation in x with roots a^3 and b^3

2. it is given that m is a constant and the sum of roots of the equation 10x^2+3mx+2=0 is greater than the product of roots by 7/10.

a) find the value of m
b) from the result of (a), solve the equation 10x^2+3mx+2=0

3. if a and b are the roots of a quadratic equation, where a+b=4 and ab=12

a) prove that the equation has no real roots.
b)find the complex roots of the equation. (express the answers as surds in their simplest form if necessary.)

4. let k be a constant. it is given that a and b are the roots of the equation x^2-kx+3=0

a) prove that a^2=ka-3
b) express a^2+kb in terms of k

另外,想請大家幫我答埋以下問題:
http://hk.knowledge.yahoo.com/question/question?qid=7010122501232

回答 (1)

不用客氣

2010-12-29 7:46 am

✔ 最佳答案

1. a)
Since a and b are the roots of x² + kx - 2 = 0,
a + b = -k
and ab = -2

a² + b² = 13
(a² + 2ab + b²) - 2ab = 13
(a + b)² - 2ab = 13
(-k)² - 2(-2) = 13
k² = 9
k = 3 or k = -3 (rejected, for k > 0)

Hence, k = 3

1. b)
a³ + b³
= (a + b)³ - 3a²b - 3ab²
= (a + b)³ - 3ab(a + b)
= (-3)³ - 3(-2)(-3)
= -45

a³b³
= (ab)³
= (-3)³
= -27

The required equation is: x² + 45x - 27 = 0
find a quadratic equation in x with roots a^3 and b^3

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2. a)
Sum of the roots = -3m/10
Product of the roots = 2/10

(-3m/10) - (2/10) = (7/10)
-3m - 2 = 7
-3m = 9
m = -3

2. b)
10x² + 3(-3)x + 2 = 0
10x² - 9x + 2 = 0
(5x - 2)(2x - 1) = 0
x = 2/5 or x = 1/2

----------
3. a)
The equation is: x² - 4x + 12 = 0

Determinant, Δ
= (-4)² - 4(1)(12)
= 16 - 48
= -32 < 0

Since Δ < 0, the equation has no real roots.

3. b)
x = {-(-4) ± √[(-4)² - 4(1)(12)]}/2
x = [4 ± (4√2)i]/2
x = 2 + (2√2)i or x = 2 - (2√2)i

----------
4. a)
Sum of the roots: a + b = k ...... (1)
Product of the roots: ab = 3 ...... (2)

From (1):
b = k - a ...... (3)

Put (3) into (2):
a(k - a) = 3
ka - a² = 3
-ka + a² = -3
a² = ka - 3

4. b)
a² + kb
= (ka - 3) + kb
= ka + kb - 3
= k(a + b) - 3
= k(k) - 3
= k² - 3

參考: micatkie

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