find the quadratic polynomial,the sum and product of whose zeroes are root 2 and -3/2 respectively.find its zeroes
Answers
Answered by
624
hey...
We know that every quadratic equation is based on this relation..
p(x) = kx² - (α+β)x + αβ
where, α and β are the zeroes of given polynomial.
Now putting values of (α+β)and αβ in above equation, we get ...
p (x) = x² - (√2)x + (-3/2) = 0
x² - √2x - 3/2 = 0
Let us multiply both sides by 2, we get...
2x² - 2√2x -3 = 0
Hence, the required polynomial is 2x² - 2√2x -3.
Hope you got it......
We know that every quadratic equation is based on this relation..
p(x) = kx² - (α+β)x + αβ
where, α and β are the zeroes of given polynomial.
Now putting values of (α+β)and αβ in above equation, we get ...
p (x) = x² - (√2)x + (-3/2) = 0
x² - √2x - 3/2 = 0
Let us multiply both sides by 2, we get...
2x² - 2√2x -3 = 0
Hence, the required polynomial is 2x² - 2√2x -3.
Hope you got it......
rahul1233:
so it is not 2p(x)?
thanks for the roots!
Yes
you are welcome
okay, i understood, thank you!
Now multiplying 2 on both sides,
We get, 2x2 - 2root2 - 3 = 0 , and we know that, P(x) = 0, So again this equation too is equal to P(X)
So P(x) doesn't change to 2P(x) !
Thanks For Co-operation
Answered by
194
Hi !!!
This is your answer...
By using formula,
kx² - (α+β)x + αβ
α and β are two zeroes of the equation ...
Putting sum of zeroes and product of zeroes in above equation.
p (x) = x² - (√2)x + (-3/2) = 0
x² - √2x - 3/2 = 0
Multiply the polynomial by 2, we get
2x² - 2√2x - 3
Hope It helps
This is your answer...
By using formula,
kx² - (α+β)x + αβ
α and β are two zeroes of the equation ...
Putting sum of zeroes and product of zeroes in above equation.
p (x) = x² - (√2)x + (-3/2) = 0
x² - √2x - 3/2 = 0
Multiply the polynomial by 2, we get
2x² - 2√2x - 3
Hope It helps
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