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F.4 A.maths Discriminant6
發問:
Discriminate the solutions of the following quadraitc equations. 1. x^2+5x+3+a=0 2. x^2+(a+b)x+ab=0 3. The quadratic equation (b-c)x^2+(c-a)x+(a-b)=0 has a repeated real solution. Prove that b=(a+c)/2 (You can rearrange b=(a+c)/2 as a+c-2b=0)
1. x^2+5x+3+a=0 Delta = 25 - 4(3+a) = 13 - 4a To have real roots , 13 - 4a>=0 , a <= 13/4 in other words , no real roots if 13- 4a<0 or a > 13/4 The roots are repeated when a= 13/4 and distinct when a < 13/4. 2. x^2+(a+b)x+ab=0 Delta = (a+b)^2 - 4ab = a^2 - 2 ab + b^2 = (a-b)^2 >= 0 Therefore, the equation has real roots and they are rational . The roots are repeated when a=b and distinct when a =/= b . 3.(b-c)x^2+(c-a)x+(a-b)=0 Since the equation admits repeated real roots for x , delta = 0 Delta = (c-a)^2 - 4 (b-c)(a-b) = 0 (c-a)^2 = 4(b-c)(a-b) = - 4[b^2 -(a+c)b +ac] 4b^2 - 4(a+c)b + (c+a)^2 = 0 which is a quadratic equation of b and [ 2b - (c+a)]^2= 0 by observation and it follows that b = (a+c)/2
其他解答:
F.4 A.maths Discriminant6
發問:
Discriminate the solutions of the following quadraitc equations. 1. x^2+5x+3+a=0 2. x^2+(a+b)x+ab=0 3. The quadratic equation (b-c)x^2+(c-a)x+(a-b)=0 has a repeated real solution. Prove that b=(a+c)/2 (You can rearrange b=(a+c)/2 as a+c-2b=0)
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最佳解答:1. x^2+5x+3+a=0 Delta = 25 - 4(3+a) = 13 - 4a To have real roots , 13 - 4a>=0 , a <= 13/4 in other words , no real roots if 13- 4a<0 or a > 13/4 The roots are repeated when a= 13/4 and distinct when a < 13/4. 2. x^2+(a+b)x+ab=0 Delta = (a+b)^2 - 4ab = a^2 - 2 ab + b^2 = (a-b)^2 >= 0 Therefore, the equation has real roots and they are rational . The roots are repeated when a=b and distinct when a =/= b . 3.(b-c)x^2+(c-a)x+(a-b)=0 Since the equation admits repeated real roots for x , delta = 0 Delta = (c-a)^2 - 4 (b-c)(a-b) = 0 (c-a)^2 = 4(b-c)(a-b) = - 4[b^2 -(a+c)b +ac] 4b^2 - 4(a+c)b + (c+a)^2 = 0 which is a quadratic equation of b and [ 2b - (c+a)]^2= 0 by observation and it follows that b = (a+c)/2
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