Question

If the sum of the cubes of zeroes of the quadratic polynomial f(t) =t^2-8t+p is 224,find the value of p. ​

Answer 1

Answer:

p = 12

Step-by-step explanation:

By Comparing "t^2 -8t + p" by Ax^2 + Bx + C

We Get

A  = 1

B = -8

C = p

Let The Zeroes of (t^2 -8t + p) be α and β

Hence

α + β = -b/a

α + β = -(-8)/1

α + β = 8

ATQ

α^3 + β^3 = 224

( α + β ) * ( α^2  + β^2 -α*β ) = 224   [ Using Identity A^3 + B^3 ]

        (8)* ( α^2  + β^2 -α*β ) = 224

                 α^2  + β^2 -α*β  = 224/8

                 α^2  + β^2 -α*β  = 28

     [(α + β) ^2 - 2α*β] - α*β  = 28 [ By Using Identity (A+B)^2 = A^2+B^2+2AB]

                        (8)^2 - 3α*β = 28

                                - 3α*β  = 28 - 64

                                - 3α*β  = - 36

                                    α*β  = (-36)/(-3)

                                    α*β  =  12 -------(1)

Also

                               α*β = C/A

                               α*β = p  -------------(2)

By Equating 1 And 2

                  we get,

                         p = 12