Solve : `(x^(2002)+10x^(2001))/(10x^(2000))=957.9`

Explanation

`(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.