Question

If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial f(x) = $3x^2-7x-6,$ find a polynomial whose zeros are ${\alpha }^2$ and ${\beta }^2$ and $2\alpha +3\beta$ and $3\alpha +2\beta$

Answer 1

Given: $f\left(x\right)=3x^2-7x-6$

Which is of the from $a{x}^2+bx+c$

$\Rightarrow a=3,b=-7,c=-6$

Now, $\alpha +\beta =-\frac{b}{a}=\frac{-(-7)}{3}$

$\alpha +\beta =\frac{7}{3}$....(1)

and $\alpha \beta =\frac{c}{a}=\frac{-6}{3}$

$\alpha \beta =-2$...(2)

(1) ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha +\beta$

$={\left(\frac{7}{3}\right)}^2-2(-2)$ [ from (1) & (2)]$

$=\frac{49}{9}+4=\frac{49+36}{9}=\frac{85}{9}$

${\alpha }^2{\beta }^2=(\alpha \beta {)}^2=(-2{)}^2=4$

The pohynomical whose roots are ${\alpha }^2,{\beta }^2$ is given by , $x^2-\left(sumofroots\right)x+productofroots$

$=x^2-\frac{85x}{9}+4$

$=\frac{9x^2-85x+36}{9}$

$=\frac{1}{9}(9x^2-85x+36)$

(2) $(2\alpha +3\beta )+(3\alpha +2\beta )=5\alpha +5\beta$

$=5(\alpha +\beta )$

$=5\cdot \frac{7}{3}$

$=\frac{35}{3}$

$(2\alpha +3\beta )+(3\alpha +2\beta )=6{\alpha }^{2}4\alpha \beta +9\alpha \beta +6{\beta }^2$

$=6({\alpha }^2+{\beta }^2)+13\alpha \beta$

$=6\left\{\right(\alpha +\beta {)}^2-2\alpha \beta \}+B\alpha \beta$

$=6(\alpha +\beta {)}^2-12\alpha \beta +B\alpha \beta$

$=6(\alpha +\beta {)}^2+\alpha \beta$

$=6{\left(\frac{7}{3}\right)}^2+(-2)$

$=6\cdot \frac{49}{9}-2$

$=\frac{98}{3}-2$

$=\frac{98-6}{3}$

$=\frac{92}{3}$

the required polynomial whose you are $(2\alpha +3\beta )$ and $(3\alpha +2\beta )$ is

$x^2-\frac{35}{3}x+\frac{92}{3}$

$=\frac{1}{3}(3x^2-35x+92)$