Question

If $\alpha ,\beta ,\gamma$ are the roots of $2x^3-5x^2-7x+8=0$, then the equation whose roots are $-\alpha ,-\beta ,-\gamma$, is:

• A. $2x^3+5x^2+7x+8=0$
• B. $2x^3+5x^2-7x-8=0$
• C. $-2x^3-5x^2+7x+8=0$
• D. $2x^3-5x^2+7x-8=0$

Answer 1

Answer: (B), (C)

The correct options are
B $2x^3+5x^2-7x-8=0$
C $-2x^3-5x^2+7x+8=0$

First, make the coefficient of the highest power of x in the given equation unity.
i.e. $x^3-\frac{5}{2x^2}-\frac{7}{2x}+4=0$

now since $\alpha$,$\beta$ ,$\gamma$ are the solutions for the given equations, thus the given equation can be written in the form:
$(x-\alpha )(x-\beta )(x-\gamma )=0$
$\Rightarrow x^3-\left(\Sigma \alpha \right)x^2+\left(\Sigma \alpha \beta \right)x-\alpha \beta \gamma =0$

Comparing the coefficients of this equation with that of the given equation gives
$\Sigma \alpha =\frac{5}{2}$ ....(1)
$\Sigma \alpha \beta =-\frac{7}{2}$ ....(2)
$\alpha \beta \gamma =-4$ ....(3)

Similarly,equation which has $-\alpha$ ,$-\beta$ ,$-\gamma$ as it's roots can be written in the form
$(x+\alpha )(x+\beta )(x+\gamma )=0$

replacing the values in the above equation from (1), (2) and (3) and solving we get the equation as,
$2x^3+5x^2-7x-8=0$