Question

The equation whose roots exceed the roots of $4x^4+32x^3+83x^2+76x+21=0$ by $2$ is:

• A. $4x^4+13x^2+9=0$
• B. $4x^4-13x^2+9=0$
• C. $4x^4+12x^2-9=0$
• D. $4x^4-13x^2-9=0$

Answer 1

The correct option is B $4x^4-13x^2+9=0$

Given: equation $4x^4+32x^3+83x^2+7x+21=0$

To find the equation whose roots exceed by 2 than the roots of the given equation

Sol: In order to increase the root by 2, we diminish the roots by -2 by dividing $4x^4+32x^3+83x^2+7x+21=0$ successively by $x+2$

The division is shown in the above figure.

Hence the required transformed equation is

$4x^4-13x^2+9=0⟹(x^2-1)(4x^2-9)=0\therefore x=\pm 1,x=\pm \frac{3}{2}$

are the roots of transformed equation.