Question

For any real number $a_3,a_4\dots a_{85}$, the roots of the equation $a_{85}{x}^{85}+a_{84}{x}^{84}+...+a_3{x}^3+{3x}^2+2x+1=0$, are not all

• A. real
• B. imaginary
• C. same
• D. cannot say

Answer 1

Answer: (A), (C)

The correct options are
A real
C same

Let $f\left(x\right)=a_{85}{x}^{85}+a_{84}{x}^{84}+...+a_3{x}^3+{3x}^2+2x+1$

Let $g\left(x\right)=f\left(\frac{1}{x}\right)$

$f\left(0\right)=1$.

Therefore, $0$ is not a root of $f\left(x\right)=0$

If $\alpha$ is a root of $f\left(x\right)=0$, then $\frac{1}{\alpha }$ is a root of $g\left(x\right)=0$

Let the roots of the two equations be ${\alpha }_1,{\alpha }_2\dots {\alpha }_i,\dots {\alpha }_{85}$ and ${\beta }_1,{\beta }_2\dots {\beta }_i,\dots {\beta }_{85}$ respectively, where ${\beta }_i={\frac{1}{\alpha }}_i$.

$g\left(x\right)=x^{85}+2x^{84}+3x^{83}+a_3{x}^{82}\dots \dots +a_{84}x+a_{85}$

$\sum {\left(\frac{1}{{\alpha }_i}\right)}^2={\left(\sum \frac{1}{\alpha }\right)}^2-2(\sum \frac{1}{{\alpha }_i{\alpha }_j})$

$\sum {\left(\frac{1}{{\alpha }_i}\right)}^2=(-2{)}^2-2\times 3=-2$

The left hand side of the equation is sum of squares of numbers.
If these numbers were all real then their sum would have been positive.
Thus all of ${\beta }_i$ cannot be real.
Hence, all of ${\alpha }_i$ cannot be real.
So, option A is true and option C follows from option A.