Transcription of MATHEMATICAL FORMULAE Algebra
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MATHEMATICAL FORMULAEA lgebra1. (a+b)2=a2+2ab+b2;a2+b2=(a+b)2 2ab2. (a b)2=a2 2ab+b2;a2+b2=(a b)2+2ab3. (a+b+c)2=a2+b2+c2+2(ab+bc+ca)4. (a+b)3=a3+b3+3ab(a+b);a3+b3=(a+b)3 3ab(a+b)5. (a b)3=a3 b3 3ab(a b);a3 b3=(a b)3+3ab(a b) b2=(a+b)(a b) b3=(a b)(a2+ab+b2) +b3=(a+b)(a2 ab+b2) bn=(a b)(an 1+an 2b+an 3b2+ +bn 1) :a:a::: :an=am+ nifm>n=1ifm=n=1an mifm<n;a2R;a6=013. (am)n=amn=(an)m14. (ab)n=an:bn15. ab n= ;a6= n=1an;an=1a Ifam=ananda6= 1;a6=0thenm=n20. Ifan=bnwheren6=0,thena= b21. Ifpx;pyare quadratic surds and ifa+px=py,thena= 0 andx=y22. Ifpx;pyare quadratic surds and ifa+px=b+pythena=bandx=y23. Ifa;m;nare positive real numbers anda6=1,thenlogamn=logam+logan24. Ifa;m;nare positive real numbers,a6=1,thenloga mn =logam logan25. Ifaandmare positive real numbers,a6=1thenlogamn=nlogam26. Ifa;bandkare positive real numbers,b6=1;k6=1,thenlogba=logkalogkb27 .
2 29. if a+ ib=0 wherei= p −1, then a= b=0 30. if a+ ib= x+ iy,wherei= p −1, then a= xand b= y 31. The roots of the quadratic equationax2+bx+c=0;a6= 0 are −b p b2 −4ac 2a The solution set of the equation is (−b+ p 2a −b− p 2a where = discriminant = b2 −4ac 32.
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