Plz answer question 5
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answer is x²-px+q. I think it's true
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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!
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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