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InterviewSolution
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Solve : `(x^(2002)+10x^(2001))/(10x^(2000))=957.9` |
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Answer» `(x^(2002)+10x^(2001))/(10x^(2000))=957.9` Multiplying both the sides by 10 `(x^(2002)+10x^(2001))/(x^(2000))=9579` `:.(x^(2002))/(x^(2000))+(10x^(2001))/(x^(2000))=9579` `:.x^(2002-2000)+10x^(2001-2000)=9579` ……… `((a^(m))/(a^(n))=a^(m-n))` `:.x^(2)+10x-9579=0` `:.x^(2)+10x+25-25-9579=0` .[Add 25 to make `x^(2)+10x` a perfect square] `:.lx^(2)+10x+25-9604=0` `:.(x+5)^(2)-(98)^(2)=0` `:.(x+5+98)` `(x+5-98)=0` `:.(x+103)(x-93)=0` `:.x+103=0` or `x-93=0` `:.x=-103` or `x=93` `93, -103` are the roots of the give quadratic equation. |
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