標題:
F.4 A.Maths
發問:
Q.1 Let f(x)=4x^2 +ax- b and g(x) =4x^2 + bx-a, where a is not equal to b. α ,β are the roots of the equation f(x) = 0 and α, γ are thr roots of the equation g(x) =0(a) Using the fact that f(α) =0 and g(α) =0, find the values of α. Hence, show that a+b = 4(b) Express β and γ in terms of a.(c) If a and b are... 顯示更多 Q.1 Let f(x)=4x^2 +ax- b and g(x) =4x^2 + bx-a, where a is not equal to b. α ,β are the roots of the equation f(x) = 0 and α, γ are thr roots of the equation g(x) =0 (a) Using the fact that f(α) =0 and g(α) =0, find the values of α. Hence, show that a+b = 4 (b) Express β and γ in terms of a. (c) If a and b are positive integers and β > γ, find the values of a, b, β and γ.
最佳解答:
Q1 (a) Since f(α) =0, 4α2 + aα - b = 0 Since g(α) = 0 4α2 + bα - a = 0 Therefore, 4α2 + aα - b = 4α2 + bα - a aα - b = bα - a (a-b)α = b-a α = -1 f(α) = 0 f(-1) = 0 4(-1)2 + a(-1) - b = 0 4 - a - b = 0 a + b = 4 (b) Sum of roots of f(x) = -a/4 α + β = -a/4 -1 + β = -a/4 β = 1 - a/4 Product of roots of g(x) = -a/4 αγ = -a/4 (-1)γ = -a/4 γ = a/4 (c) Since β＞γ, 1 - a/4＞a/4 4 - a＞a 2a＜4 a＜2 Since a is positive integer, a = 1.【1 is the only positive ＜2】 So, β = 1 - a/4 = 1 - 1/4 = 3/4 γ = a/4 = 1/4 Also, a + b = 4 1 + b = 4 b = 3 In conclusion, a = 1, b = 3, β = 3/4, γ = 1/4. Hope it helps! ^^
其他解答:
F.4 A.Maths
發問:
Q.1 Let f(x)=4x^2 +ax- b and g(x) =4x^2 + bx-a, where a is not equal to b. α ,β are the roots of the equation f(x) = 0 and α, γ are thr roots of the equation g(x) =0(a) Using the fact that f(α) =0 and g(α) =0, find the values of α. Hence, show that a+b = 4(b) Express β and γ in terms of a.(c) If a and b are... 顯示更多 Q.1 Let f(x)=4x^2 +ax- b and g(x) =4x^2 + bx-a, where a is not equal to b. α ,β are the roots of the equation f(x) = 0 and α, γ are thr roots of the equation g(x) =0 (a) Using the fact that f(α) =0 and g(α) =0, find the values of α. Hence, show that a+b = 4 (b) Express β and γ in terms of a. (c) If a and b are positive integers and β > γ, find the values of a, b, β and γ.
最佳解答:
Q1 (a) Since f(α) =0, 4α2 + aα - b = 0 Since g(α) = 0 4α2 + bα - a = 0 Therefore, 4α2 + aα - b = 4α2 + bα - a aα - b = bα - a (a-b)α = b-a α = -1 f(α) = 0 f(-1) = 0 4(-1)2 + a(-1) - b = 0 4 - a - b = 0 a + b = 4 (b) Sum of roots of f(x) = -a/4 α + β = -a/4 -1 + β = -a/4 β = 1 - a/4 Product of roots of g(x) = -a/4 αγ = -a/4 (-1)γ = -a/4 γ = a/4 (c) Since β＞γ, 1 - a/4＞a/4 4 - a＞a 2a＜4 a＜2 Since a is positive integer, a = 1.【1 is the only positive ＜2】 So, β = 1 - a/4 = 1 - 1/4 = 3/4 γ = a/4 = 1/4 Also, a + b = 4 1 + b = 4 b = 3 In conclusion, a = 1, b = 3, β = 3/4, γ = 1/4. Hope it helps! ^^
其他解答:
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