Question

α, β are zeroes of the polynomial p(x) = 3x² – 6x – 5. Find the value of `1/α^2 + 1/β^2`.

Answer 1

For p(x) = 3x² – 6x – 5

a = 3, b = –6, c = –5

The relationship between zeroes and coefficients of a quadratic polynomial ax² + bx + c is:

Sum of zeroes `(α + β) = - b/a`

`α + β = - (-6)/3 = 2`

Product of zeroes `(αβ) = c/a`

`αβ = (-5)/3`

Now, find α² + β²:

`α^2 + β^2 = (2)^2 - 2 ((-5)/3)`

= `4 + 10/3`

= `(12 + 10)/3`

= `22/3`

Now, calculate the required expression:

`1/α^2 + 1/β^2 = (α^2 + β^2)/(αβ)^2`

= `(22/3)/(- 5/3)^2`

= `(22/3)/(25/9)`

= `22/3 xx 9/25`

= `(22 xx 3)/25`

= `66/25`