Question

If a, ß are the zeros of the polynomial x² + 6x + 2 then (1/a+1/B​

Answer 1

Answer:

-3

Step-by-step explanation:

x² + 6x + 2 =0

[x= -b +(or)- √(b²- 4ac)]

by applying above formula, we can get

x = (-6 + √28)/2 or x = (-6 -√28)/2

1/a+1/b=[2(-6-√28)+2(-6+√25)]/36-28

=2(-6-6-√28+√28)/8

= -12/4

= -3

Answer 2

GIVEN :-

• ɑ and β are zeros of the polynomial x² + 6x + 2.

TO FIND :-

• The value of 1/ɑ + 1/β.

SOLUTION :-

★ x² + 6x + 2 = 0

As we know that,

⇒ Sum of zeroes = - coefficient of x/coefficient of x².

⇒ Sum of zeroes = -b/a.

⇒ ɑ + β = -b/a.

⇒ ɑ + β = -6/1

⇒ ɑ + β = -6

Similarly, we will find the product of zeros.

⇒ product of zeros = constant term/coefficient of x².

⇒ product of zeros = c/a.

⇒ ɑ × β = c/a

⇒ ɑ × β = 2/1

⇒ ɑ × β = 2.

Now we have to find the value of 1/ɑ + 1/β,

⇒ 1/ɑ + 1/β.

Taking L.C.M as ɑ × β.

⇒ 1/ɑ + 1/β = (ɑ + β)/(ɑ × β)

⇒ 1/ɑ + 1/β = -6/2

⇒ 1/ɑ + 1/β = -3

Hence required value of 1/ɑ + 1/β = is -3.