Two Three Two Three Two Three, And?

Algebra Level 4

${ 2x }^{ 2001 }+{ 3x }^{ 2000 }+{ 2x }^{ 1999 }+{ 3x }^{ 1998 }+\ldots +2x+3=0$

Find the number of real root(s) of the equation above.

The answer is 1.

2 solutions

Sai Venkatesh
Dec 11, 2014

The left hand side of the given equation can be rewritten as

$(2x+3)(x^{2000}+x^{1998}+ \ldots +x^{2}+1) = 0.$

From this, we see that one real root is $x = -\frac { 3 }{ 2 }.$

However, the remaining factor of $x^{2000}+x^{1998}+..........+x^{2}+1$ is a sum of squares plus one, which means this factor is always positive. Therefore, there is only one real root.

i didn't understand the last part...plz explain clearly

Md Ashiqur rahman - 5 years, 6 months ago

If we take second part of equation

That is x^2000 + ..... +x+1=0

x^2000.......+x=-1

Therefore the roots are imaginary imaginary

Prakhar Dhumas - 5 years, 4 months ago

Hi Ashiq, All of the x's in the expression are powers by even positive integers, so whatever the value of x is the the value must be positive or at least zero (if x = 0) and the last term of the expression is 1, so the expression value is greater than equal to 1, whatever the value of the x is. This prove the claim that (2x+3) = 0 → x = -(3/2)

Mahabubul Islam - 5 years, 1 month ago

Николай Нагибин
Nov 30, 2015

Obnivious that x=1 and x=-1 isn't roots. The left hand of equation is sum of 2×(x^2001+x^2000+x+...+x+1) + (x^2000+x^1998+...+x^2). Each of () is sum of geometric progression, and 2×((x^2002-1)/(x-1) + (x^2002-1)/(x^2-1)=0

(2(x+1)+1)×(x^2002-1)=0 X=-1.5 or x=1 or x=-1 But two of last roots isn't roots (see before). Then, there is one root x=-1.5 P.s. I'm sorry for my bad english!

Two Three Two Three Two Three, And?

The answer is 1.

2 solutions

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