Question

if alpha and beta are the zeros of x*2 + 7 x + 12, then find the value of 1/ alpha + 1/beta + 2 alpha beta
who is answering first will be marked as brainelist please fast

Answer 1

Hello,
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ɑ and β are the zeroes of the x² + 7x + 12

Now,

x² + 7x + 12 = 0

=> x² + 3x + 4x + 12 = 0

=> x(x + 3) + 4(x + 3) = 0

=> (x + 3)(x + 4) = 0

So,

x + 3 = 0

=> x = - 3

Again,

x + 4 = 0

=> x = - 4

ɑ = - 3 and β = - 4

So,

1/ɑ + 1/β + 2ɑβ

= 1/-3 + 1/-4 + 2 × (-3) × (-4)

= -1/3 - 1/4 + 24

= (-4 - 3)/12 + 24

= -7/12 + 24

= (-7 + 288)/12

= 281/12


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ArchitectSethRollins

Answer 2

Given Equation is x^2 + 7x + 12.

Here a = 1, b = 7, c = 12.

Given α,β are the zeroes of equation.

(i)

We know that Sum of zeroes = -b/a

⇒ α + β = -7/1

⇒ α + β = -7

(ii)

We know that product of zeroes = c/a

⇒ αβ = 12/1

⇒ αβ = 12.

Now,

⇒ (1/α) + (1/β) + 2αβ

⇒ [(α + β)/αβ] + 2αβ

⇒ [(-7)/12] + 2(12)

⇒ -7/12 + 24

⇒ 281/12

Hope it helps!