Root distributivity

Algebra Level 2

Let x 1 , x 2 , x 3 , … , x n ( n ∈ N ) x_1, x_2, x_3, \ldots , x_n (n \in \mathbb{N}) be real, non-negative numbers which satisfy the following equation:

x 1 + x 2 + x 3 + … + x n = x 1 + x 2 + x 3 + … + x n \sqrt{x_1 + x_2 + x_3 + \ldots + x_n} = \sqrt{x_1} + \sqrt{x_2} + \sqrt{x_3} + \ldots + \sqrt{x_n}

How many of these numbers, at most, can be different than 0 0 ?


The answer is 1.

Novak Radivojević by Novak Radivojević , 23, Serbia

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2 solutions

Squaring the given equation we get

∑ i , j , i < j x i x j = 0 \displaystyle \sum_{i, j, i<j} x_ix_j=0 . So, at most one term can be different from zero, since otherwise we will be left with an invalid equality, equating a non-zero term with zero. For instance, say x i ≠ 0 , x j ≠ 0 , i ≠ j x_i\neq 0,x_j\neq 0,i\neq j . All the other terms are zero. Then the above result will yield x i x j = 0 x_ix_j=0 , which is impossible.

1 Helpful 1 Interesting 1 Brilliant 0 Confused

Apart from 0 0 , 1 1 is the only that satisifes this equation as if x 1 , . . . , x n ≥ 2 x_1, ..., x_n \geq 2 , the R.H.S results in an finite number of square roots or the L.H.S becomes either a surd or it doesn't match the R.H.S

Therefore, the answer is 1 \fbox 1 .

0 Helpful 0 Interesting 0 Brilliant 1 Confused

if x 1 , … , x 2 ≥ 2 x_1,\ldots,x_2\geq2 , the R.H.S results in an finite number of square roots or the L.H.S becomes either a surd or it doesn't match the R.H.S

So what if the premise is true? Why must the conclusion follow?

Pi Han Goh - 11 months, 3 weeks ago

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I tried it on paper - that is what I got (albeit in words).

A Former Brilliant Member - 11 months, 3 weeks ago

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