Question

if alpha and beta are the zeroes of p(x)=3x²+2x-6 then find the value of alpha²+beta²​

Answer 1

Answer:

p(x) =3x2 + 2x -6

α+β =2/3

αβ =-6/3= -2

α2+β2= (α+β)2 +2αβ

α2+β2=(2/3)2 + 2(-2)

α2+β2 =4/9 -4

α2+β2= -32/9

Answer 2

3x² + 2x - 6 = 0

$\mathfrak\red{sum\:of\:zeroes\::-}$

$- \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

$- \frac{2}{3}$

$\mathfrak\blue{product\:of\:zeroes\::-}$

$\frac{constant \: term}{coefficient \: {x}^{2} }$

$- \frac{6}{3}$

$- 2$

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

${ ( \frac{ - 2}{3} )}^{2} - 2( \frac{ - 2}{ \: 3} )( - 2)$

$\frac{4}{9} - \frac{8}{3}$

$\frac{4 - 24}{9}$

$- \frac{ 20}{9}$

$<marquee> please mark as brainliest.........$