Question

If α, β are zeroes of the polynomial p(x) = 5x² – 7x – 3, then form quadratic polynomial whose zeroes are `2/α` and `2/β`.

Answer 1

p(x) = 5x² – 7x – 3

Given that, α and β are the zeroes of p(x)

∴ `α + β = - ((-7))/5`

⇒ `α + β = 7/5`

And `αβ = (-3)/5`

Now, Zeroes of required polynomial are `2/α` and `2/β`

So, sum of zeroes = `2/α + 2/β`

= `2(1/α + 1/β)`

= `2((β + α)/(αβ))`

= `2((7/5)/((-3)/5))`

= `2(7/(-3))`

= `(-14)/3`

⇒ Sum of zeroes = `2/α xx 2/β = 4/(αβ)`

And product of zeroes = `4/((-3)/5) = (-20)/3`

Now, required polynomial is given by

f(n) = K [x² – (Sum of zeroes)x + Product of zeroes] {where K ≠ 0}

⇒ `f(n) = K [x^2 - ((-14)/3)x + ((-20)/3)]`

⇒ `f(n) = K [x^2 + 14/3x - 20/3]`

⇒ `f(n) = K [(3x^2 + 14x - 20)/3]`

If K = 3, then f(x) = 3x² + 14x – 20