Math, asked by priya7125choudhury, 10 months ago

If alpha and beta are the zeros of the Quarditic polynomial p(x) = x2-x-4 then find 1/ alpha + 1/beta - alpha beta

Answers

Answered by EthicalElite
1

Sol : We have quadratic equation x² - x - 4.

Given α and ß are their zeroes.

We know that,

Sum of roots = - ( coefficient of x )/ coefficient of x²

α + ß = - ( - 1 ) / 1

α + ß = 1 / 1 = 1.

Now,

Product of roots = constant term / coefficient of x²

αß = ( - 4 ) / 1

αß = -4.

1. 1/α + 1/ß - αß

     ß + α

=  -------------  - αß

         αß

By substituting the values of ( α + ß ) and ( αß ),

= ( 1 / -4 ) - ( - 4 )

= ( - 1 / 4 ) + 4

      - 1 + 16

=  ----------------

           4

= 15 / 4.

2. α/ß + ß/α + 2 ( 1/α + 1/ß ) + 3αß

    α² + ß² + 2ß + 2α

=  ---------------------------   + 3αß

             αß

      α² + ß² + 2 ( α+ß )

=  -------------------------  + 3αß   ----- eq.1

            αß

Now ,we don't have the value of ( α² + ß² ), so let's find it ,

( α + ß )² = α² + ß² + 2 αß

By substituting the values of ( α + ß ) and αß in above equation,

 ( 1 )² = α² + ß² + 2 ( - 4 )

 1 = α² + ß² - 8 

 α² + ß² = 1 + 8

 α² + ß² = 9

Now by substituting the values of ( α² + ß² ) ,αß and ( α + ß ) in eq.1,

    9 + 2 ( 1 )

= --------------- + 3 ( - 4 )

        -4

     9 + 2

=  --------------  - 12

       -4 

      - 11

=  --------------  - 12

        4

    -11 - 48

= --------------

         4

= -59/4.

 

Hope it helps you,

Please mark me as brainlist.

Answered by Anonymous
16

Answer:

Sol : We have quadratic equation x² - x - 4.

Given α and ß are their zeroes

We know that,

Sum of roots = - ( coefficient of x )/ coefficient of x²

α + ß = - ( - 1 ) / 1

α + ß = 1 / 1 = 1.

Now

Product of roots = constant term / coefficient of x²

αß = ( - 4 ) / 1

αß = -4.

1. 1/α + 1/ß - αß

ß + α

= ------------- - αß

By substituting the values of ( α + ß ) and ( αß ),

= ( 1 / -4 ) - ( - 4 )

= ( - 1 / 4 ) + 4

- 1 + 16

= ----------------

4

= 15 / 4.

2. α/ß + ß/α + 2 ( 1/α + 1/ß ) + 3αß

α² + ß² + 2ß + 2α

= --------------------------- + 3αß

αß

α² + ß² + 2 ( α+ß )

= ------------------------- + 3αß ----- eq.1

αß

Now ,we don't have the value of ( α² + ß² ), so let's find it ,

( α + ß )² = α² + ß² + 2 αß

By substituting the values of ( α + ß ) and αß in above equation,

( 1 )² = α² + ß² + 2 ( - 4 )

1 = α² + ß² - 8

α² + ß² = 1 + 8

α² + ß² = 9

Now by substituting the values of ( α² + ß² ) ,αß and ( α + ß ) in eq.1,

9 + 2 ( 1 )

= --------------- + 3 ( - 4 )

-4

9 + 2

= -------------- - 12

-4

- 11 -48

4

=-59/4

Similar questions