Expression with Two Terms

Algebra Level 3

If n n is a positive integer, how many real roots can the expression below have?

x 2 n − x n \large x^{2n}-x^{n}

2 n 2n n n 3 3 or 2 2 3 3 2 2 1 1

Kaizen Cyrus by Kaizen Cyrus , 18, Philippines

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1 solution

Chew-Seong Cheong
Feb 15, 2019

To find the roots, we have

x 2 n − x n = 0 x n ( x n − 1 ) = 0 \begin{aligned} x^{2n} - x^n & = 0 \\ x^n\left(x^n-1\right) & = 0 \end{aligned}

⟹ { x n = 0 ⟹ x = 0 for all n x n − 1 = 0 ⟹ { x = 1 for all n x = − 1 if n is even \implies \begin{cases} x^n = 0 & \implies x = 0 \quad \text{for all }n \\ x^n - 1 = 0 & \implies \begin{cases} x = 1 & \text{for all }n \\ x = - 1 & \text{if }n \text{ is even} \end{cases} \end{cases}

Therefore, there are { 2 solutions { x = 0 x = 1 if n is odd. 3 solutions { x = − 1 x = 0 x = 1 if n is even. \begin{cases} \text{2 solutions } \begin{cases} x = 0 \\ x = 1 \end{cases} & \text{if }n \text{ is odd.} \\ \text{3 solutions }\begin{cases} x = -1 \\ x = 0 \\ x = 1 \end{cases} & \text{if }n \text{ is even.} \end{cases}

So, the answer is 2 or 3 \boxed{\text{2 or 3}} depending on if n n is odd or even respectively.

1 Helpful 0 Interesting 2 Brilliant 0 Confused

All true but the question is not stated right. It says how many it can have, that's essentially asking for a maximum amount of roots. That would be 3. I even wondered if the question was tricking me into chosing 2 or 3.

Peter van der Linden - 2 years, 3 months ago

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