Question

If $\alpha ,\beta ,\gamma ,\sigma$ are the roots of the equation $x^4+4x^3-6x^2+7x-9=0$, then the value of $(1-{\alpha }^2)(1-{\beta }^2)(1-{\gamma }^2)(1-{\sigma }^2)$, is:

• A. $-75$
• B. $25$
• C. $0$
• D. $75$

Answer 1

The correct option is D $75$

Given equation is $x^4+4x^3-6x^2+7x-9=0$

$(x-\alpha )(x-\beta )(x-\gamma )(x-\sigma )=x^4+4x^3-6x^2+7x-9$

and if roots are $-\alpha ,-\beta ,-\gamma ,-\sigma$, then equation would be

$(x+\alpha )(x+\beta )(x+\gamma )(x+\sigma )=x^4-4x^3-6x^2-7x-9$

Substitute $x=1$ in both equations, we get

$(1-\alpha )(1-\beta )(1-\gamma )(1-\sigma )=-3(1+\alpha )(1+\beta )(1+\gamma )(1+\sigma )=-25$

So, $(1-{\alpha }^2)(1-{\beta }^2)(1-{\gamma }^2)(1-{\sigma }^2)=75$