標題:
A.Maths (Quadratic Equation)
發問:
Solver the problems: 1)Prove that (a^2+b^2)x^2+2(a+b)x+2=0 has no real roots if a and b are unequal 2)Prove that if a,b and c are real nos,then the equations (x-a)(x-b)=c^2 always has real roots.
最佳解答:
Δ= [ 2 ( a + b )]^2 - 4 ( a^2 + b^2 )( 2 ) = 4 ( a + b )^2 - 8a^2 - 8b^2 = 4a^2 + 8ab + 4b^2 - 8a^2 - 8b^2 = - 4a^2 + 8ab - 4b^2 = - 4 ( a^2 - 2ab + b^2 ) = - 4 ( a - b )^2 0 So the equation always has real roots. 2007-10-05 22:11:23 補充： Sorry for a typing error: the last line should be ( a - b )^2 + 4c^2 >= 0
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其他解答:
1)Prove that (a^2+b^2)x^2+2(a+b)x+2=0 has no real roots if a and b are unequal (a^2+b^2)x^2+2(a+b)x+2=0 delta = 4(a+b)^2 - 4(a^2+b^2)(2) = 4 [ (a+b)^2 - 2(a^2+b^2) ] = 4 [ a^2 + 2ab + b^2 - 2a^2 - 2b^2 ] = 4 [ -a^2 + 2ab - b^2] = -4 [ a^2 - 2ab + b^2] = -4 ( a - b)^2 = 0 so, (x-a)(x-b)=c^2 always has real roots 2007-10-05 21:54:43 補充： 大眼睇過龍, sorry, 第二條應該係delta = (a + b)^2 - 4( ab - c^2 )= (a + b)^2 - 4ab + 4 c^2= a^2 + 2ab + b^2 - 4ab + 4 c^2= a^2 - 2ab + b^2 + 4 c^2= (a - b)^2 + 4 c^2 >= 0so, (x-a)(x-b)=c^2 always has real roots上面位人兄都做錯少少, 請支持我一票, 唔該
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