Question

If alpha and betha are the roots of the equation 2x^2 - 6x + 18 = 0, then the value of alpha^4 + bheta^4 is

Answer 1

Answer:

- 81           (complex roots)

Step-by-step explanation:

For quadratic equations in form of ax^2 + bx + c = 0,

Sum of roots = -b/a

Product of roots = c/a

      In the given equation 2x^2  - 6x + 18 = 0,

        a = 2    ; b = -6       ; c = 18

∴ if α and β are the roots

    ⇒ α + β = - (-6)/2 = 3         ...(1)

    ⇒ αβ = 18/2 = 9

Square on both sides of (1):

     ⇒ α^2 + β^2 + 2αβ = 9

     ⇒ α^2 + β^2 = 9 - 2(9)

     ⇒ α^2 + β^2 = -9

Square on both sides again:

         ⇒ α^4 + β^4 + 2α^2β^2 = 81

         ⇒ α^4 + β^4 + 2(αβ)^2 = 81

         ⇒ α^4 + β^4 = 81 - 2(81)

         ⇒ α^4 + β^4 = -81