Plz solve this...plz plz plz plz plzI want with Solution...

1.	Plz solve this...plz plz plz plz plzI want with Solution...
Answer» Solution :- given quadratic equation is 9x² + 6X + 1 = 0 . so, → sum of roots = (ɑ + β) = -b/a = (-6/9) = (-2/3) ----- Eqn.(1) and, → product of roots = ɑ•β = c/a = 1/9 ---- Eqn.(2) now, roots of other equation are 1/ɑ and 1/β . so, → sum of roots = 1/ɑ + 1/β = {(ɑ + β) / ɑ•β} putting VALUES from Eqn.(1) and Eqn.(2) , → sum of roots = (-2/3) / (1/9) = (-2/3) * 9 = (-6) and, → product of roots = 1/ɑ * 1/β = 1/(ɑ•β) = 1/(1/9) = 9 then, Required equation will be, → x² - sum of roots * X + product of roots = 0 → x² - (-6)x + 9 = 0 → x² + 6x + 9 = 0 (Ans.) LEARN more :- JEE mains Question :- brainly.in/question/22246812 . FIND all the zeroes of the polynomial x4 – 5x3 + 2x2+10x-8, if two of its zeroes are 4 and 1. brainly.in/question/39026698

Answer»

Solution :-

given quadratic equation is 9x² + 6X + 1 = 0 .

so,

→ sum of roots = (ɑ + β) = -b/a = (-6/9) = (-2/3) ----- Eqn.(1)

and,

→ product of roots = ɑ•β = c/a = 1/9 ---- Eqn.(2)

now, roots of other equation are 1/ɑ and 1/β .

so,

→ sum of roots = 1/ɑ + 1/β = {(ɑ + β) / ɑ•β}

putting VALUES from Eqn.(1) and Eqn.(2) ,

→ sum of roots = (-2/3) / (1/9) = (-2/3) * 9 = (-6)

and,

→ product of roots = 1/ɑ * 1/β = 1/(ɑ•β) = 1/(1/9) = 9

then, Required equation will be,

→ x² - sum of roots * X + product of roots = 0

→ x² - (-6)x + 9 = 0

→ x² + 6x + 9 = 0 (Ans.)

LEARN more :-

JEE mains Question :-

. FIND all the zeroes of the polynomial x4

– 5x3 + 2x2+10x-8, if two of its zeroes are 4 and 1.

Discussion