2 0 0 ( x + 2 0 ) ( x + 1 0 ) − 1 0 0 ( x + 1 0 ) ( x + 3 0 ) + 2 0 0 ( x + 3 0 ) ( x + 2 0 ) = 1
Find out the number of distinct root(s) of the equation above.
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It would be clearer to explain that:
1. Shift 1 over to the LHS. The expression is a polynomial of degree at most 2. It is either a quadratic polynomial, a linear polynomial, a non-zero constant, or zero.
2. Since there are 3 distinct roots, hence the expression must be zero. Thus there are infinitely many solutions.
(This is exactly what you are saying, but the phrasing could be improved on).
Not a level 4 problem.
If we have 2 0 0 ( x + 2 0 ) ( x + 1 0 ) − 1 0 0 ( x + 1 0 ) ( x + 3 0 ) + 2 0 0 ( x + 3 0 ) ( x + 2 0 ) = 1 , then can simplify to:
2 0 0 ( x + 2 0 ) ( x + 1 0 ) − 1 0 0 ( x + 1 0 ) ( x + 3 0 ) + 2 0 0 ( x + 3 0 ) ( x + 2 0 ) = 1 ;
or ( x + 2 0 ) ( x + 1 0 ) − 2 ( x + 1 0 ) ( x + 3 0 ) + ( x + 3 0 ) ( x + 2 0 ) = 2 0 0 ;
or ( x 2 + 3 0 x + 2 0 0 ) − ( 2 x 2 + 8 0 x + 6 0 0 ) + ( x 2 + 5 0 x + 6 0 0 ) = 2 0 0 ;
or ( 2 x 2 − 2 x 2 ) + ( 8 0 x − 8 0 x ) + ( 2 0 0 − 2 0 0 + 6 0 0 − 6 0 0 ) = 0 ;
or 0 = 0 .
Hence, there is an infinite number of distinct roots that solves this equation.
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By observing the equation it is clear that its greatest power of x is two, so by the fundamental theorem of algebra it must have at least two roots(i am saying at least because we have not yet opened brackets) Now put x=-10,-20 and -30 one by one.....equation get satisfied by all these values.but by the fundamantel theoram of algebra it can not have more then two roots provided that equation is not an identity....but as proved it is satisfied by 3 distinct values of x hence it must be an identity.....so there are infinite many roots