Not So Big!

Algebra Level 3

( x b ) ( x c ) ( x d ) ( a b ) ( a c ) ( a d ) + ( x c ) ( x d ) ( x a ) ( b c ) ( b d ) ( b a ) + ( x d ) ( x a ) ( x b ) ( c d ) ( c a ) ( c b ) + ( x a ) ( x b ) ( x c ) ( d a ) ( d b ) ( d c ) = ? \begin{aligned} & \dfrac{(x-b)(x-c)(x-d)}{(a-b)(a-c)(a-d)} \\+ & \dfrac{(x-c)(x-d)(x-a)}{(b-c)(b-d)(b-a)} \\ +& \dfrac{(x-d)(x-a)(x-b)}{(c-d)(c-a)(c-b)} \\ +& \dfrac{(x-a)(x-b)(x-c)}{(d-a)(d-b)(d-c)} \end{aligned} ~=~?

Hint : An equation of n th n^ \text{th} degree with more than n n roots is an identity. This is studied extensively in the wiki: The method of undetermined coefficients .


The answer is 1.

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3 solutions

Rishabh Jain
Mar 23, 2016

Let f ( x ) f(x) denote the given expression such that degree of f ( x ) f(x) is 3 3 .

Then we can see f ( x ) = 1 f(x)=1 for x = a , b , c , d x=a,b,c,d or alternatively f ( x ) 1 = 0 f(x)-1=0 has four roots.

An n t h nth degree equation can have more than n n roots only if it's an identity !!

Therefore f ( x ) 1 = 0 x R f(x)-1=0~\forall x\in\mathfrak R or f ( x ) = 1 x R \boxed{f(x)=1}~\forall x\in\mathfrak R .

Ha ha ,nice use of that hint.

Rohit Udaiwal - 5 years, 2 months ago
Yoga Nugraha
Mar 24, 2016

Let a = x a = x then go on ;)

Is it a way to solve these questions or it just worked this time

Chaitnya Shrivastava - 5 years, 2 months ago

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Because the question does not state any value of x (or any of the variables), and we know the answer is an integer, we can substitute any number in.

A P - 5 years, 2 months ago

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Very well!Thank you

Chaitnya Shrivastava - 5 years, 2 months ago
Rohit Nair
Mar 25, 2016

Since the answer is not a variable, simply put x=a, b, c or d. And get the answer in seconds!!!!

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