Not the Fundamental Theorem of Calculus

Algebra Level 4

( x + 20 ) ( x + 10 ) 200 ( x + 10 ) ( x + 30 ) 100 + ( x + 30 ) ( x + 20 ) 200 = 1 \frac{(x+20)(x+10)}{200}-\frac{(x+10)(x+30)}{100}+\frac{(x+30)(x+20)}{200}=1

Find out the number of distinct root(s) of the equation above.

Infinite many 0 3 2

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2 solutions

Aman Sharma
Oct 19, 2014

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

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).

Calvin Lin Staff - 6 years, 2 months ago

Not a level 4 problem.

Alex Zhong - 6 years, 2 months ago
Tom Engelsman
Apr 2, 2017

If we have ( x + 20 ) ( x + 10 ) 200 ( x + 10 ) ( x + 30 ) 100 + ( x + 30 ) ( x + 20 ) 200 = 1 \frac{(x+20)(x+10)}{200} - \frac{(x+10)(x+30)}{100} + \frac{(x+30)(x+20)}{200} = 1 , then can simplify to:

( x + 20 ) ( x + 10 ) 200 ( x + 10 ) ( x + 30 ) 100 + ( x + 30 ) ( x + 20 ) 200 = 1 ; \frac{(x+20)(x+10)}{200} - \frac{(x+10)(x+30)}{100} + \frac{(x+30)(x+20)}{200} = 1;

or ( x + 20 ) ( x + 10 ) 2 ( x + 10 ) ( x + 30 ) + ( x + 30 ) ( x + 20 ) = 200 ; (x+20)(x+10) - 2(x+10)(x+30) + (x+30)(x+20) = 200;

or ( x 2 + 30 x + 200 ) ( 2 x 2 + 80 x + 600 ) + ( x 2 + 50 x + 600 ) = 200 ; (x^2 + 30x + 200) - (2x^2 +80x + 600) + (x^2 + 50x + 600) = 200;

or ( 2 x 2 2 x 2 ) + ( 80 x 80 x ) + ( 200 200 + 600 600 ) = 0 ; (2x^2 - 2x^2) + (80x - 80x) + (200 - 200 + 600 - 600) = 0;

or 0 = 0. 0 = 0.

Hence, there is an infinite number of distinct roots that solves this equation.

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