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標題:
F4 Amaths (Quadratic equations)
發問:
Given that a , b amd c are the lengths of the sides of triangle ABC. Prove that if (c+a)x^2 + 2bx + (c-a) = 0 has two equal roots, then triangle ABC is a right-angled triangle.
最佳解答:
(c + a)x2 + 2bx + (c - a) = 0 The equation has two equal roots So, discriminant = 0 i.e. (2b)2 – 4(c + a)(c - a) = 0 4b2 – 4(c2 – a2) = 0 b2 – c2 + a2 = 0 a2 + b2 = c2 Since a, b and c are the lengths of the sides of triangle ABC. By pythagras theorem, Triangle ABC is a right-angled triangle.
其他解答:
F4 Amaths (Quadratic equations)
發問:
Given that a , b amd c are the lengths of the sides of triangle ABC. Prove that if (c+a)x^2 + 2bx + (c-a) = 0 has two equal roots, then triangle ABC is a right-angled triangle.
最佳解答:
(c + a)x2 + 2bx + (c - a) = 0 The equation has two equal roots So, discriminant = 0 i.e. (2b)2 – 4(c + a)(c - a) = 0 4b2 – 4(c2 – a2) = 0 b2 – c2 + a2 = 0 a2 + b2 = c2 Since a, b and c are the lengths of the sides of triangle ABC. By pythagras theorem, Triangle ABC is a right-angled triangle.
其他解答:
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Let the angle between the side a and side b is the right angle Therefore, c^2 = a^2 + b^2 (c+a)x^2 + 2bx + (c-a) = 0 If the equation has two equal roots, discriminant = 0 discriminant = (2b)^2 - 4 (c+a)(c-a) 4b^2 + 4a^2 - 4c^2 = 4(a^2+b^2 - c^2) = 0 => there are two equal root for the equation
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