Relations and Functions

Relations and Functions

Question

Solve each of the following equations:

 space space space space space space space space space space straight x squared space plus space straight x space plus space fraction numerator 1 over denominator square root of 2 end fraction equals space 0

Answer

Here,     space space space space space space space space space space space space straight D space equals space space straight b squared minus 4 ac equals space left parenthesis 1 right parenthesis squared minus 4 left parenthesis 1 right parenthesis open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses equals space 1 space minus fraction numerator 4 over denominator square root of 2 end fraction. fraction numerator square root of 2 over denominator square root of 2 end fraction equals space 1 space minus space 2 space square root of 2 space space less than space 0
Therefor, roots of given equation are imaginary and are given by

space space space space space space space space space space space space space straight x space equals space fraction numerator negative straight b space plus-or-minus space square root of straight D over denominator 2 straight a end fraction comma space straight x space equals space fraction numerator negative 1 space plus-or-minus space square root of 1 space minus space 2 square root of 2 end root over denominator 2 left parenthesis 1 right parenthesis end fraction
i.e.  space space space space space space space space space space straight x space equals space fraction numerator negative 1 space plus space square root of 2 square root of 2 space minus space 1 space straight i end root end root over denominator 2 end fraction comma space fraction numerator negative 1 space minus square root of 2 square root of 2 space minus space 1 space straight i end root end root over denominator 2 end fraction

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