Question

If $2,5,7,-4$ are the roots of $a{x}^4+b{x}^3+c{x}^2+dx+e=0$ then the roots of $a{x}^4-b{x}^3+c{x}^2-dx+e=0$ are:

• A. $2,5,7,-4$
• B. $-2,-5,-7,4$
• C. $2,5,7,4$
• D. $2,-5,7,-4$

Answer 1

The correct option is B $-2,-5,-7,4$

The equation $a{x}^4+b{x}^3+c{x}^2+dx+3=0$ will be converted to $a{x}^4-b{x}^3+c{x}^2-dx+3=0$ if we substitute $-x$ in place of $x$ in the frist equation. So, if $2,5,7,-4$ are the roots of the first equation then $-2,-5,-7,4$ will be the roots of the second equation.