Question

If ɑ and ʙ are the zeroes of the polynomial x² + 12x + 4 then the value of ( 1/ɑ + 1/ʙ ) is what?
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Answer 1

Answer:

x²+12x+4

x=-12±rootb²-4ac/2a

x=-12±root 144-16/2

solve this u get urs answer perfectly

Answer 2

Given : -

If α and β are the zeroes of the polynomial x² + 12x + 4 = 0 .

Required to find : -

• find the value of ( 1/α + 1/β ) ?

Solution : -

Quadratic equation : x² + 12x + 4 = 0

α and β are the zeroes of the polynomial .

Here we can solve this question using 2 methods !

The standard form of the Quadratic equation is

ax² + bx + c = 0

On comparing the standard form of a quadratic equation with the given polynomial

Here,

• a = 1

• b = 12

• c = 4

1st method

We know that ;

There is a relationship between the zeroes of the Quadratic equation with respective to coefficients of the quadratic equation .

So,

The relation between the sum of the zeroes and the coefficients is ;

α + β = - coefficient of x/coefficient of x²

α + β = - b/a

α + β = -(12)/1

α + β = - 12

Similarly,

The relation between the product of the zeroes and the coefficients is ;

α.β = constant term/coefficient of x²

α.β = c/a

α.β = 4/1

α.β = 4

Now,

Let's find the value of ( 1/α + 1/β )

1/α + 1/β =

α + β/α.β

substituting the values ;

- 12/4

-3

2nd method

The Quadratic formula is ;

$\boxed{ \sf{x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }}$

Here,

• a = 1

• b = 12

• c = 4

Substituting these in the formula ;

$\sf x = \dfrac{ - ( 12 ) \pm \sqrt{ ( 12 )^2 - 4 ( 1 ) ( 4)} }{ 2 ( 1 )} \\ \\ \\ \sf x = \dfrac{ - 12 \pm \sqrt{ ( 12 )^2 - 4 ( 1 ) ( 4)} }{ 2 } \\ \\ \\ \sf x = \dfrac{ - 12 \pm \sqrt{ 144 - 16} }{ 2 } \\ \\ \\ \sf x = \dfrac{ - 12 + \sqrt{128}}{2 } \quad ( and ) \quad x = \dfrac{ - 12 - \sqrt{128}}{2} \\ \\ \\ \implies \sf \alpha = \dfrac{ - 12 + \sqrt{128}}{2 } \quad ( and ) \quad \beta = \dfrac{ - 12 - \sqrt{128}}{2}$

Now,

Let's find the value of 1/α + 1/β !

$\sf \to \dfrac{1}{ \alpha} + \dfrac{1}{ \beta} \\ \\ \to \sf \dfrac{ \alpha + \beta }{ \alpha \beta } \\ \\ \to \sf \dfrac{ \dfrac{-12 + \sqrt{128}}{2} + \dfrac{- 12 - \sqrt{128}}{2} }{ \dfrac{- 12 + \sqrt{128}}{2} \times \dfrac{- 12 - \sqrt{ 128}}{2} } \\ \\ \\ \to \sf \dfrac{ \dfrac{- 12 + \sqrt{128} - 12 - \sqrt{128}}{2} }{\dfrac{( - 12 )^2 - ( \sqrt{128} )^2}{4} } \\ \\ \\ \to \sf \dfrac{ \dfrac{ - 12 - 12}{2 } }{ \dfrac{144 - 128}{4} } \\ \\ \\ \to \sf \dfrac{ \dfrac{- 24}{2} }{ \dfrac{16}{4} } \\ \\ \to \sf \frac{ - 24 }{ \dfrac{16}{2} } \\ \\ \to \sf \dfrac{ - 48 }{16} \\ \\ \implies \sf - 3$

Therefore,

Value of 1/α + 1/β = - 3