Question

if alpha and beta are the zeroes of the polynomial 3x^2+2x-6,find the values of
1)alpa-beta
2)alpha^2+beta^2
3)1/alpha+1/beta

Answer 1

$(\alpha-\beta)^2=\alpha^2-2\alpha\beta+\beta^2\\(\alpha+\beta)^2=\alpha^2+2\alpha\beta+\beta^2=\alpha^2+2\alpha\beta-4\alpha\beta+\beta^2+4\alpha\beta=\\\alpha^2-2\alpha\beta+\beta^2+4\alpha\beta=(\alpha-\beta)^2+4\alpha\beta\\(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\\\alpha-\beta=\pm\sqrt{(\alpha+\beta)^2-4\alpha\beta}\\\\(\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta\\\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\\\\\frac1\alpha+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta}$

On basis of Vieta's formulas:
$f(x)=ax^2+bx+c=3x^2+2x-6\\\alpha+\beta=\frac{-b}a=\frac{-2}3\\\alpha\beta=\frac{c}a=-2$

$1)\ \alpha-\beta=\pm\sqrt{(\alpha+\beta)^2-4\alpha\beta}=\pm\sqrt{(\frac{-2}3)^2-4(-2)}=\pm\frac23\sqrt{19}\\\\2)\ \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=(\frac{-2}3)^2-2(-2)=4\frac49\\\\3){<span>\frac{1}{\alpha}}+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta}=\frac{\frac{-2}3}{-2}=\frac13$