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a.maths-quadratic equation (F.4) urgent
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1a) x2+4x-1+k(x+1)2=0 x2+4x-1+k(x2+2x+1)=0 x2+4x-1+kx2+2kx+k=0 (k+1)x2+(4+2k)x+(k-1)=0 i)For real and distinct roots,△ > 0, so (4+2k)2-4(k+1)(k-1) > 0 (2+k)2-(k2-1) > 0 4+4k+k2-k2+1 > 0 4k > -5 k > - 5 / 4 ii)For equal roots, △ = 0 , so (4+2k)2-4(k+1)(k-1) = 0 4+4k+k2-k2+1 = 0 k = - 5 / 4 b)From x2+ax-5=0, A+B=-a, AB=-5 From x2-19x+b=0, A2+B2=19, A2B2=b So, b=(-5)2 = 25 (A+B)2-2AB=19 (-a)2-2(-5)=19 a2=9 a=3 or –3 2) x2+2(b+c-a)x+2bc=a2 x2+2(b+c-a)x+(2bc-a2)=0 △ = 4(b+c-a)2-4(1)(2bc-a2) =4 [(b+c)-a]2-8bc+4a2 =4(b+c)2-8(b+c)(a)+4a2-8bc+4a2 =4(b2+2bc+c2)-8ab-8ac+8a2-8bc =4b2+8bc+4c2-8ab-8ac+8a2-8bc =4b2+4c2-8ab-8ac+8a2 =4c2-8ac+4a2+4a2-8ab+4b2 =4(c-a)2+4(a-b)2≧0 Hence the equation has real roots for all real values of a,b and c.
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a.maths-quadratic equation (F.4) urgent
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1.(a) Find the possible values of k where k ≠-1, for which the roots of the quadratic equation x^2+4x-1+k(x+1)^2=0 are (i) real and distinct, (ii) equal(b) A and B are the roots of the quadratic equation x^2+ax-5=0 where a is a constant. If A^2 and B^2 are the roots of the quadratic equation x^2-19x+b=0 where b is... 顯示更多 1.(a) Find the possible values of k where k ≠-1, for which the roots of the quadratic equation x^2+4x-1+k(x+1)^2=0 are (i) real and distinct, (ii) equal (b) A and B are the roots of the quadratic equation x^2+ax-5=0 where a is a constant. If A^2 and B^2 are the roots of the quadratic equation x^2-19x+b=0 where b is a constant, find the possible values of a and b. 2. Show that the quadratic equation x^2+2(b+c-a)x+2bc=a^2 has real roots for all real values of a, b and c.最佳解答:
1a) x2+4x-1+k(x+1)2=0 x2+4x-1+k(x2+2x+1)=0 x2+4x-1+kx2+2kx+k=0 (k+1)x2+(4+2k)x+(k-1)=0 i)For real and distinct roots,△ > 0, so (4+2k)2-4(k+1)(k-1) > 0 (2+k)2-(k2-1) > 0 4+4k+k2-k2+1 > 0 4k > -5 k > - 5 / 4 ii)For equal roots, △ = 0 , so (4+2k)2-4(k+1)(k-1) = 0 4+4k+k2-k2+1 = 0 k = - 5 / 4 b)From x2+ax-5=0, A+B=-a, AB=-5 From x2-19x+b=0, A2+B2=19, A2B2=b So, b=(-5)2 = 25 (A+B)2-2AB=19 (-a)2-2(-5)=19 a2=9 a=3 or –3 2) x2+2(b+c-a)x+2bc=a2 x2+2(b+c-a)x+(2bc-a2)=0 △ = 4(b+c-a)2-4(1)(2bc-a2) =4 [(b+c)-a]2-8bc+4a2 =4(b+c)2-8(b+c)(a)+4a2-8bc+4a2 =4(b2+2bc+c2)-8ab-8ac+8a2-8bc =4b2+8bc+4c2-8ab-8ac+8a2-8bc =4b2+4c2-8ab-8ac+8a2 =4c2-8ac+4a2+4a2-8ab+4b2 =4(c-a)2+4(a-b)2≧0 Hence the equation has real roots for all real values of a,b and c.
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