Math, asked by nisargcbhavsarcsn, 1 year ago

Plz answer question 5

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Answered by jaisandeep
0
answer is x²-px+q. I think it's true
Answered by siddhartharao77
0
Note: I am writing Alpha and Beta as a and b.

Given f(x) = x^2 + px + q.

It is in the form of ax^2 + bx + c, we get a = 1, b = p, c = q

Now,

= > We know that sum of zeroes = -b/a

a + b = -p/1

a + b = -p.


= > We know that product of zeroes = c/a

ab = q/1

ab = q.

Now, We have Zeroes of a required Quadratic polynomial as 1/a, 1/b.

Sum of roots of the quadratic polynomial:

= > (1/a + 1/b)

= > (a + b)/ab

= > -p/q.


The product of roots of Quadratic polynomial:

= > 1/a * 1/b

= > 1/ab

= > 1/q.


Therefore the required polynomial is:

= > x^2 - (sum of zeroes)x + product of zeroes) 

=  >x^2 - (-p/q)x + (1/q)

= > qx^2 + px + 1



Hope this helps!

siddhartharao77: :-)
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