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F.4 AM
發問:
Consider the equation x^2+(2+m)x+2m-1=0, where m is a constant.a)Find the discriminant of the equation.b)Show that the equation always has two unequal real roots.c)When the discriminant is at the minimum, (c1)find the value of m ,(c11)solve the equation.d)Let a and b be the roots of equation when m-4,... 顯示更多 Consider the equation x^2+(2+m)x+2m-1=0, where m is a constant. a)Find the discriminant of the equation. b)Show that the equation always has two unequal real roots. c)When the discriminant is at the minimum, (c1)find the value of m , (c11)solve the equation. d)Let a and b be the roots of equation when m-4, where a>b. (d1)find a and b in surd form (d11)find the exact value of a^5-b^5
最佳解答:
Consider the equation x^2+(2+m)x+2m-1=0, where m is a constant. a)Find the discriminant of the equation. Ans: Δ = b^2 -4ac =(2+m)^2-4(2m-1) =4+4m+m^2-8m+4 =m^2-4m+8// b)Show that the equation always has two unequal real roots. Δ =m^2-4m+8 =m^2-4m+4+4 =(m-2)^2+4 >0 (since any intiger ^2 > 0)// c)When the discriminant is at the minimum (c i )find the value of m Ans: the minimum value is 2 (cii)solve the equation. x^2+(2+m)x+2m-1=0 =>x^2+(2+2)x+2(2)-1=0 =>x^2+4+3=0 =>(x+3)(x+1)=0 =>x = -1 or -3//
F.4 AM
發問:
Consider the equation x^2+(2+m)x+2m-1=0, where m is a constant.a)Find the discriminant of the equation.b)Show that the equation always has two unequal real roots.c)When the discriminant is at the minimum, (c1)find the value of m ,(c11)solve the equation.d)Let a and b be the roots of equation when m-4,... 顯示更多 Consider the equation x^2+(2+m)x+2m-1=0, where m is a constant. a)Find the discriminant of the equation. b)Show that the equation always has two unequal real roots. c)When the discriminant is at the minimum, (c1)find the value of m , (c11)solve the equation. d)Let a and b be the roots of equation when m-4, where a>b. (d1)find a and b in surd form (d11)find the exact value of a^5-b^5
最佳解答:
Consider the equation x^2+(2+m)x+2m-1=0, where m is a constant. a)Find the discriminant of the equation. Ans: Δ = b^2 -4ac =(2+m)^2-4(2m-1) =4+4m+m^2-8m+4 =m^2-4m+8// b)Show that the equation always has two unequal real roots. Δ =m^2-4m+8 =m^2-4m+4+4 =(m-2)^2+4 >0 (since any intiger ^2 > 0)// c)When the discriminant is at the minimum (c i )find the value of m Ans: the minimum value is 2 (cii)solve the equation. x^2+(2+m)x+2m-1=0 =>x^2+(2+2)x+2(2)-1=0 =>x^2+4+3=0 =>(x+3)(x+1)=0 =>x = -1 or -3//
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