Question

if alpha and beta are the zeroes of the polynomial 3x2 +8x +2 find the value of (i) alpha 2 + beta 2 (ii) alpha2 - beta 2

Answer 1

$\bigstar$ Given :

• α and β are the zeroes of the polynomial 3x²+8x+2

$\bigstar$ To Find :

• (i) α² + β²
• (ii) α² - β²

$\bigstar$ Solution :

Compare given equation 3x²+8x+2 with ax²+bx+c , we get ,

• a = 3 , b = 8 , c = 2

$\bold{Sum\;of\;zeroes,\alpha +\beta =\frac{-b}{a} }\\\\\implies \bold{\alpha +\beta =\frac{-8}{3} }$

$\bold{Product\;of\;zeroes\;,\alpha \beta =\frac{c}{a} }\\\\\implies \bold{\alpha \beta =\frac{2}{3} }$

( i )

$\bold{\alpha ^2+\beta ^2}\\\\\implies \bold{(\alpha+\beta )^2-2.\alpha\beta }\\\\\implies \bold{(\frac{-8}{3} )^2-2(\frac{2}{3} )}\\\\\implies \bold{\frac{64}{9}-\frac{4}{3} }\\\\\implies \bold{\frac{(64-12)}{9} }\\\\\implies \bold{\frac{52}{9} }$

_________________

$(\alpha +\beta )^2=(\alpha -\beta )^2+4\alpha \beta\\\\\implies (\frac{-8}{3} )^2=(\alpha-\beta )^2+4(\frac{2}{3} )\\\\\implies \frac{64}{9}=(\alpha-\beta ) ^2+\frac{8}{3} \\\\\implies (\alpha -\beta )^2=\frac{64}{9}-\frac{8}{3} \\\\\implies (\alpha -\beta )^2=\frac{40}{9} \\\\\implies (\alpha -\beta )=\frac{2\sqrt{10} }{3}$

( ii )

$\bold{\alpha^2-\beta^2 }\\\\\implies \bold{(\alpha+\beta )(\alpha-\beta )}\\\\\implies \bold{(\frac{-8}{3} )(\frac{2\sqrt{10} }{3} )}\\\\\implies \bold{\frac{-16\sqrt{10} }{9} }$