Question

Find the equation whose roots are equal in magnitude but opposite in sign to the roots of the equation $x^5$- 3 $x^3$ + 2 $x^2$ + 4x + 1 = 0.

• A. $x^5$- 3$x^3$ + 2$x^2$ + 4x + 1 = 0
• B. $x^5$ + 2$x^3$ + 3$x^2$ + 5x - 1 = 0
• C. $x^5$+ $x^3$ + $x^2$ + x + 1 = 0
• D. $x^5$ - 3$x^3$ - 2$x^2$ + 4x - 1 = 0

Answer 1

The correct option is D

$x^5$ - 3$x^3$ - 2$x^2$ + 4x - 1 = 0

Solution: $x^5$- 3 $x^3$ + 2 $x^2$ + 4x + 1 = 0 ____ (1)

Let x be a root of the given equation and y a root of the required equation.

Then y = - x or x = - y

Put x = - y in the equation 1, we have

- $y^5$ + 3 $y^3$ + 2 $y^2$ - 4y + 1 = 0

Express it in terms of x

- $x^5$ + 3 $x^3$ + 2 $x^2$ - 4x + 1 = 0

$x^5$ - 3 $x^3$ - 2 $x^2$ + 4x - 1 = 0