標題:
發問:
1. without calculator, show √3 + 1 is one of the root of x^3 - 4x^2 + 2x + 4 = 0. 2. Given ( 1 + x + 3x^2 )^8=1 + 8x + 52x^2 + 224x^3. Hence evaluate (1.0103)^8 correct to 4 sig. fig. 3. In (1 + mx)^n, where n, m are +ve integers, the coefficient of x^2 is twice the coeff. of x. Find the values of m and n.
最佳解答:
1. without calculator, show √3 + 1 is one of the root of x^3 - 4x^2 + 2x + 4 = 0. (√3 + 1)^3 - 4(√3 + 1)^2 + 2(√3 + 1) + 4 = (√3)^3 + 3(√3)^2 + 3(√3) + 1 - 4(3 + 2√3 + 1) + 2√3 + 2 + 4 = 3√3 + 9 + 3√3 + 1 - 4(3 + 2√3 + 1) + 2√3 + 2 + 4 = 0 2. Given ( 1 + x + 3x^2 )^8=1 + 8x + 52x^2 + 224x^3. Hence evaluate (1.0103)^8 correct to 4 sig. fig. put x = 0.01 (1.0103)^8 = ( 1 + x + 3x^2 )^8 = 1 + 8x + 52x^2 + 224x^3 = 1 + 8(0.01) + 52(0.01)^2 + 224(0.01)^3 = 1 + 8(0.01) + 52(0.0001) + 224(0.000001) = 1 + 0.08 + 0.0052 + 0.000224 = 1.085 3. In (1 + mx)^n, where n, m are +ve integers, the coefficient of x^2 is twice the coeff. of x. Find the values of m and n. (1 + mx)^n = 1 + mnx + [n(n-1)/2] (mx)^2 + ... = 1 + mnx + [n(n-1)m^2 /2] x^2 + ... n(n-1)m^2 /2 = 2mn (n-1)m /2 = 2 (n-1)m = 4 m = 1, 2 , 3 or 4 when m = 1, n = 5 when m = 2, n = 3 when m = 3, rejected when m = 4, n = 2 (m,n) = (1,5), (2,3) or (4,2)
其他解答:FAD2A23AB937987B
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A.maths binomial發問:
1. without calculator, show √3 + 1 is one of the root of x^3 - 4x^2 + 2x + 4 = 0. 2. Given ( 1 + x + 3x^2 )^8=1 + 8x + 52x^2 + 224x^3. Hence evaluate (1.0103)^8 correct to 4 sig. fig. 3. In (1 + mx)^n, where n, m are +ve integers, the coefficient of x^2 is twice the coeff. of x. Find the values of m and n.
最佳解答:
1. without calculator, show √3 + 1 is one of the root of x^3 - 4x^2 + 2x + 4 = 0. (√3 + 1)^3 - 4(√3 + 1)^2 + 2(√3 + 1) + 4 = (√3)^3 + 3(√3)^2 + 3(√3) + 1 - 4(3 + 2√3 + 1) + 2√3 + 2 + 4 = 3√3 + 9 + 3√3 + 1 - 4(3 + 2√3 + 1) + 2√3 + 2 + 4 = 0 2. Given ( 1 + x + 3x^2 )^8=1 + 8x + 52x^2 + 224x^3. Hence evaluate (1.0103)^8 correct to 4 sig. fig. put x = 0.01 (1.0103)^8 = ( 1 + x + 3x^2 )^8 = 1 + 8x + 52x^2 + 224x^3 = 1 + 8(0.01) + 52(0.01)^2 + 224(0.01)^3 = 1 + 8(0.01) + 52(0.0001) + 224(0.000001) = 1 + 0.08 + 0.0052 + 0.000224 = 1.085 3. In (1 + mx)^n, where n, m are +ve integers, the coefficient of x^2 is twice the coeff. of x. Find the values of m and n. (1 + mx)^n = 1 + mnx + [n(n-1)/2] (mx)^2 + ... = 1 + mnx + [n(n-1)m^2 /2] x^2 + ... n(n-1)m^2 /2 = 2mn (n-1)m /2 = 2 (n-1)m = 4 m = 1, 2 , 3 or 4 when m = 1, n = 5 when m = 2, n = 3 when m = 3, rejected when m = 4, n = 2 (m,n) = (1,5), (2,3) or (4,2)
其他解答:FAD2A23AB937987B
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