Question

if alpha and beta are the roots of 3x2+5x-7=0 then 1÷(3alpha +5)2 +1÷(3beta+5)2

Answer 1

alfa(a)+beta(b)= -b/a= -5/3
alfa×beta=c/a= -7/3
So,
1/(3a+5)^2 + 1/(3b+5)^2
= 1/(9a^2+25+30a) + 1/(9b^2+25+30b)
=(9b^2+25+30b+9a^2+25+30a)
_________________________
(9a^2+25+30a)(9b^2+25+30b)

=9(a^2+b^2)+50+30(a+b)
____________________
((3a+5)(3b+5))^2

=9((a+b)^2-2ab)+50+30(a+b)
______________________
(9ab+15(a+b)+25)^2

=9(25/9+14/3)+50-30×5/3
_____________________
(9×(-7/3)-15×5/3+25)^2

=67
_____
(-21)^2

=67/441

PLZ MARK IT AS BRAINLIEST

Answer 2

alpha and beta are the roots of
3x² +5-7 = 0

alpha.beta = -7/3
3alpha.beta = -7

alpha + beta = -5/3
3alpha + 3beta = - 5
3alpha +5 = -3beta

in the same way ,
3beta + 5 = -3alpha

now ,
1/(3alpha +5)² + 1/(3beta +5)²

=1/(-3beta)² + 1/( -3alpha)²

={9(alpha² + beta²}/81alpha².beta²

={(3alpha +3beta)²-2.3(-7)}/9(3alpha.beta)²

={ (-5)²+42}/9(-7)²

=67/9×49

=67/441