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If at-by-cz and b2-ac, prove that2xz |
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Answer» A^x= b^y = c^z , from this if we like to express A & c in terms of b —- A^x = b^y, => A = b^(y/x) , similarly , as c^z = b^y Therefore , c = b^(y/z). It is also given that —— b^2 = Ac, or,b^2 = {b^(y/x)}*{b^(y/z)} ( putting the value of A & c defined in terms of b ) Or, b^2= b^(y/x+y/z) or, b^2 = b^{( yz +xy)/xz } or, (xy + yz)/xz =2 (as base on the both side equals (b) , power is equal) or, xy + yz = 2xz or, y(x+z) = 2xz So,y = 2xz/(x+z) |
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