Question

If α and β are zeroes of the polynomial t²- t - 4, form a quadratic polynomial whose zeroes are 1/α and 1/β .

Answer 1

Let p(x) = t²  -t -4

On comparing with ax²+bx+c
a= 1,  b= -1 , c= -4

Given, α and β are the zeroes of p(x).
Sum of zeroes (α+ β) = -b/a
α+ β = - (-1)/1
α+ β = 1………………..(1)

Product of zeroes (α. β) = c/a
α. β = -4/1= -4

Given, zeroes are  1/α and 1/β .
Sum of zeroes=  (1/α+ 1/β) =( α+ β )/α β
Sum of zeroes= 1/ -4= -1/4
[From equation 1]

Product of zeroes= 1/α. 1/β = 1/(α.β)
= 1/-4= -¼
[From equation 2]

Required Polynomial= t²-(Sum of zeroes)t +( Product of zeroes)
= t² -(-1/4)t + (-¼)
= t² + t/4 -¼
= 4t² +t -1
Hence,the Required Polynomial is  4t² +t -1

HOPE THIS WILL HELP YOU....

Answer 2

Polynomials,

We have,
A polynomial who's zeros are alpha(a) and beta(b).

Let the polynomial having the zeros 1/a and 1/b as g(t)

Now let, f(t) = t²- t - 4

As we know that the sum of the zeros of a polynomial = -(coefficient of x)/(coefficient of x²)

So (a + b) = -(-1)/1 = 1.........(1)

And product of the zeros = (constant term)/coefficient of x²)
So, ab = (-4)/1 = -4........(2)

The polynomial having zeros 1/a and 1/b are given by,
k{x²- (sum of the zeros)x + product of the zeros. [Where k is any constant term]

So,
g(t) = t² - (1/a + 1/b)t + 1/ab
= t²- {(a + b)/ab}t + 1/ab
= t²- 1/(-4) + 1/(-4) [substituting the values from equation 1 and 2]
= t² + 1/4t -1/4
= 4t²+ t - 1

Therefore the desired polynomial is 4t²+ t - 1

That's it
Hope it helped (･ิω･ิ)