Question

If $\alpha +\beta =3$ and ${\alpha }^3+{\beta }^3=9$, find the quadratic equation whose roots are $\alpha$ or $\beta$ :

• A. $x^2-3x+3=0$
• B. $x+\frac{2}{x}+3=0$
• C. $x^2-2x+3=0$
• D. $x+\frac{2}{x}=3$
• E. $x^2-3x+2=0$

Answer 1

The correct option is E $x^2-3x+2=0$

Given that $\alpha +\beta =3$ and ${\alpha }^3+{\beta }^3=9$

By squaring $\alpha +\beta =3$ on both sides , we get ${\alpha }^2+{\beta }^2+2\alpha \beta =9$

$\Rightarrow {\alpha }^2+{\beta }^2=9-2\alpha \beta$

We know that ${\alpha }^3+{\beta }^3=(\alpha +\beta )({\alpha }^2+{\beta }^2-\alpha \beta )=9$

$\Rightarrow (\alpha +\beta )(9-3\alpha \beta )=9$

$\Rightarrow 3\alpha \beta =9-3=6$

$\Rightarrow \alpha \beta =2$

Therefore we have $\alpha +\beta =3$ and $\alpha \beta =2$

So the required equation is $x^2-3x+2=0$