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10分 [急] F4 A.Maths (英文)
1.Given that P and Q are the roots of the equation x^2+(k+1)x+(k-1)=0,where k is a constant.1.a.Show that P and Q are distinct real roots.1.b.Express P^3 + Q^3 in term of k.Hence,find a quadratic equation whose roots are P^3 and Q^3 in terms of k.1.c.Suppose P and Q are the roots of the quadratic equation x^2... 顯示更多 1.Given that P and Q are the roots of the equation x^2+(k+1)x+(k-1)=0,where k is a constant. 1.a.Show that P and Q are distinct real roots. 1.b.Express P^3 + Q^3 in term of k.Hence,find a quadratic equation whose roots are P^3 and Q^3 in terms of k. 1.c.Suppose P and Q are the roots of the quadratic equation x^2 +3x +1 =0. 1.c.1.By using (b), or otherwise, find a quadratic equation whose roots are P^3 and Q^3 1.c.2.Hence,find the value of (-9+4開方5)^(1/3)+(-9-4開方5)^(1/3)
最佳解答:
1.Given that P and Q are the roots of the equation x2+(k+1)x+(k-1)=0,where k is a constant. 1.a.Show that P and Q are distinct real roots. As P and Q are roots of x2 + (k+1)x + (k-1) = 0, Discriminant △ = (k+1)2 - 4(1)(k-1) = k2 + 2k + 1 - 4k + 4 = k2 - 2k + 5 = k2 - 2k + 1 + 4 = (k-1)2 + 4 > 0 As △ > 0 So P and Q are real and distinct. ======================================= 1.b.Express P3 + Q3 in term of k.Hence,find a quadratic equation whose roots are P3 and Q3 in terms of k. As P and Q are roots of x2 + (k+1)x + (k-1) = 0, Sum of roots = -(k+1)/1 P + Q = -(k-1) ... (1) Product of roots = (k-1)/1 PQ = k-1 ... (2) P3 + Q3 = (P+Q)(P2 - PQ + Q2) 【Using identity a3+b3=(a+b)(a2-ab+b2)】 = (P+Q)[(P2 + 2PQ + Q2) - 3PQ] = (P+Q)[(P+Q)2 - 3(PQ)] = [-(k+1)]【[-(k+1)]2 - 3(k-1)】 【Using (1), (2)】 = -(k+1)[k2 + 2k + 1 - 3k + 3] = -(k+1)(k2 - k + 4) ... (3) Consider P3Q3 = (PQ)3 = (k-1)3 【From (2)】 ... (4) From (3), (4), the equation with roots P3 and Q3 is x2 - [-(k+1)(k2 -k + 4)]x + (k-1)3 = 0 x2 + (k+1)(k2 - k + 4)x + (k-1)3 = 0 is the required equation ======================================= 1.c.Suppose P and Q are the roots of the quadratic equation x2 +3x +1 =0. 1.c.1.By using (b), or otherwise, find a quadratic equation whose roots are P3 and Q3 As P and Q are roots of equation x2 + 3x + 1 = 0 Putting k = 2, P and Q are roots of equation of x2 + (k+1)x + (k-1) = 0 By (b), the equation of roots P3 and Q3 is x2 + (k+1)(k2 - k + 4)x + (k-1)3 = 0 x2 + (2+1)(22 - 2 + 4)x + (2-1)3 = 0 【Putting k = 2】 x2 + 18x + 1 = 0 is the required equation ======================================= 1.c.2.Hence,find the value of (-9+4√5)1/3+(-9-4√5)1/3 From (b), the equation of roots P3 and Q3 is x2 + 18x + 1 = 0 The roots = 【-18 ±√[182 - 4(1)(1)]】/2(1) = [-18 ±√320]/2 = [-18 ±8√5]/2 = -9 ± 4√5 So P3 and Q3 are -9 + 4√5 and -9 - 4√5 respectively. As P and Q are the roots of x2 + 3x + 1 = 0 So (-9+4√5)1/3 and (-9-4√5)1/3 are the roots of x2 + 3x + 1 = 0 (-9+4√5)1/3 + (-9-4√5)1/3 = sum of roots of x2 + 3x + 1 = 0 = -3/1 = -3 So (-9+4√5)1/3 + (-9-4√5)1/3 = -3
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10分 [急] F4 A.Maths (英文)
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發問:1.Given that P and Q are the roots of the equation x^2+(k+1)x+(k-1)=0,where k is a constant.1.a.Show that P and Q are distinct real roots.1.b.Express P^3 + Q^3 in term of k.Hence,find a quadratic equation whose roots are P^3 and Q^3 in terms of k.1.c.Suppose P and Q are the roots of the quadratic equation x^2... 顯示更多 1.Given that P and Q are the roots of the equation x^2+(k+1)x+(k-1)=0,where k is a constant. 1.a.Show that P and Q are distinct real roots. 1.b.Express P^3 + Q^3 in term of k.Hence,find a quadratic equation whose roots are P^3 and Q^3 in terms of k. 1.c.Suppose P and Q are the roots of the quadratic equation x^2 +3x +1 =0. 1.c.1.By using (b), or otherwise, find a quadratic equation whose roots are P^3 and Q^3 1.c.2.Hence,find the value of (-9+4開方5)^(1/3)+(-9-4開方5)^(1/3)
最佳解答:
1.Given that P and Q are the roots of the equation x2+(k+1)x+(k-1)=0,where k is a constant. 1.a.Show that P and Q are distinct real roots. As P and Q are roots of x2 + (k+1)x + (k-1) = 0, Discriminant △ = (k+1)2 - 4(1)(k-1) = k2 + 2k + 1 - 4k + 4 = k2 - 2k + 5 = k2 - 2k + 1 + 4 = (k-1)2 + 4 > 0 As △ > 0 So P and Q are real and distinct. ======================================= 1.b.Express P3 + Q3 in term of k.Hence,find a quadratic equation whose roots are P3 and Q3 in terms of k. As P and Q are roots of x2 + (k+1)x + (k-1) = 0, Sum of roots = -(k+1)/1 P + Q = -(k-1) ... (1) Product of roots = (k-1)/1 PQ = k-1 ... (2) P3 + Q3 = (P+Q)(P2 - PQ + Q2) 【Using identity a3+b3=(a+b)(a2-ab+b2)】 = (P+Q)[(P2 + 2PQ + Q2) - 3PQ] = (P+Q)[(P+Q)2 - 3(PQ)] = [-(k+1)]【[-(k+1)]2 - 3(k-1)】 【Using (1), (2)】 = -(k+1)[k2 + 2k + 1 - 3k + 3] = -(k+1)(k2 - k + 4) ... (3) Consider P3Q3 = (PQ)3 = (k-1)3 【From (2)】 ... (4) From (3), (4), the equation with roots P3 and Q3 is x2 - [-(k+1)(k2 -k + 4)]x + (k-1)3 = 0 x2 + (k+1)(k2 - k + 4)x + (k-1)3 = 0 is the required equation ======================================= 1.c.Suppose P and Q are the roots of the quadratic equation x2 +3x +1 =0. 1.c.1.By using (b), or otherwise, find a quadratic equation whose roots are P3 and Q3 As P and Q are roots of equation x2 + 3x + 1 = 0 Putting k = 2, P and Q are roots of equation of x2 + (k+1)x + (k-1) = 0 By (b), the equation of roots P3 and Q3 is x2 + (k+1)(k2 - k + 4)x + (k-1)3 = 0 x2 + (2+1)(22 - 2 + 4)x + (2-1)3 = 0 【Putting k = 2】 x2 + 18x + 1 = 0 is the required equation ======================================= 1.c.2.Hence,find the value of (-9+4√5)1/3+(-9-4√5)1/3 From (b), the equation of roots P3 and Q3 is x2 + 18x + 1 = 0 The roots = 【-18 ±√[182 - 4(1)(1)]】/2(1) = [-18 ±√320]/2 = [-18 ±8√5]/2 = -9 ± 4√5 So P3 and Q3 are -9 + 4√5 and -9 - 4√5 respectively. As P and Q are the roots of x2 + 3x + 1 = 0 So (-9+4√5)1/3 and (-9-4√5)1/3 are the roots of x2 + 3x + 1 = 0 (-9+4√5)1/3 + (-9-4√5)1/3 = sum of roots of x2 + 3x + 1 = 0 = -3/1 = -3 So (-9+4√5)1/3 + (-9-4√5)1/3 = -3
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