Question

- If (x - 2) = 5 , find the values of
a) (x² -1/x2) (x +-1/x4)​

Answer 1

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

         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.

Answer 2

Answer:

Factors of x

2

−3x+2 are (x−1) and (x−2).

Let f(x)=x

4

−px

2

+q

Since, f(x) is divisible by (x−1) and (x−2)

∴f(1)=0 and f(2)=0

⇒f(1)=1

4

−p(1)

2

+q=0

and f(2)=2

4

−p(2)

2

+q=0

1−p+q=0....(i)

and 16−4p+q=0....(ii)

(i)−(ii), we get

−15+3p=0⇒p=5

putting value of p in eq. (i), we get q=4

Thus, we get

p=5 and q=4.