標題:
a maths(Quadratic Equations)
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發問:
the quadratic equation x^2 log a+(x+1)log b=0, where a,b are two positive constants , has two non-zero equal roots. Express b in terms of a.
最佳解答:
x2 log a+(x+1)log b=0 i.e. x2loga + xlogb +logb=0 The quadratic equation has two non-zero equal roots, ∴△=0 (logb)2-4(loga)(logb)=0 logb(logb-4loga)=0 4loga=logb a4=b a=b1/4// 2008-09-03 22:59:14 補充： a^4=b b=a^4
其他解答:
the quadratic equation x^2 log a+(x+1)log b=0, where a,b are two positive constants , has two non-zero equal roots. Express b in terms of a. Δ=(log b)^2 -4(log a)(log b)=0 log b -4(log a)=0 log b -log a^4=0 log b =log a^4 b = a^4|||||the root is (-log b)/(2 log a) =/= 0 ==> log b =/= 0 Δ=(log b)^2 - 4(log a)(log b)=0 Since log b =/= 0, log b can be divided on both sides. log b - 4 log a = 0 log b = log a^4 b = a^4
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