Question

If $\alpha ,\beta ,\gamma \text{and}\delta$ are the roots of the polynomial equation $(x^2-3x+4)(x^2+2x+5)=0$, then the quadratic equation whose roots are $a+\beta +\gamma +\delta$ and $\alpha \beta \gamma \delta$ is

• A. $x^2-x+20=0$
• B. $x^2-5x+20=0$
• C. $x^2+x-20=0$
• D. $x^2-x-10=0$

Answer 1

The correct option is A $x^2-x+20=0$

$(x^2-3x+4)(x^2+2x+5)$

Let $\alpha ,\beta$ be roots of 1st quadratic and $\gamma ,\delta$ be roots of 2nd quadratic

$\alpha +\beta =\frac{(-)(-3)}{1};\gamma +\delta =\frac{-2}{1};\alpha +\beta +\gamma +\delta =1=(-\frac{b}{a})$

$\alpha \beta =\frac{4}{1};\gamma \delta =\frac{5}{1};\alpha \beta \gamma \delta =20=\left(\frac{d}{a}\right)$

$\therefore$ equation will be $x^2-x+20=0$