Question

if alpha, beta are the roots of 9 X square + 6 X + 1 is equal to zero then equation with the roots one by Alpha One by beta is​

Answer 1

Answer:

Step-by-step explanation:

Hi friend

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Your answer

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Equation => 9x² + 6x + 1

Factorising the equation , we get ,

9x² + 6x + 1

=> 9x² + 3x + 3x + 1

=> 3x(3x + 1) + 1(3x + 1)

=> (3x + 1)(3x + 1)

Now,

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zeroes of the equation , are

3x + 1 = 0

=> 3x = - 1

=> x = - 1/3

As , the factors are same (3x + 1) , that means the zeroes are same .

Let alpha = -1/3 and beta = -1/3

Then,

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Alpha + 1/beta

\alpha  +  \frac{1}{ \beta }  \\  \\  =  >  \frac{ - 1}{3}  +  \frac{1}{ \frac{ - 1}{3} }  \\  \\  =  >  \frac{ - 1}{3}   - 3 \\  \\  =  >  \frac{ - 1 - 9}{3}  \\  \\  =  >  \frac{ - 10}{3}

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Answer 2

Answer:

Step-by-step explanation:

Equation => 9x² + 6x + 1

Factorising the equation , we get ,

9x² + 6x + 1

=> 9x² + 3x + 3x + 1

=> 3x(3x + 1) + 1(3x + 1)

=> (3x + 1)(3x + 1)

Now,

---------

zeroes of the equation , are

3x + 1 = 0

=> 3x = - 1

=> x = - 1/3

As , the factors are same (3x + 1) , that means the zeroes are same .

Let alpha = -1/3 and beta = -1/3

Then,

----------

Alpha + 1/beta

\alpha  +  \frac{1}{ \beta }  \\  \\  =  >  \frac{ - 1}{3}  +  \frac{1}{ \frac{ - 1}{3} }  \\  \\  =  >  \frac{ - 1}{3}   - 3 \\  \\  =  >  \frac{ - 1 - 9}{3}  \\  \\  =  >  \frac{ - 10}{3}

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