Write a quadratic polynomial whose zero's are -9 and 1/9?
Answers
Answer:
zeroes --> -9 & 1/9
sum of zeroes= -9+1/9
=(-81+1)/9
=-80/9
product of zeroes=-9(1/9)
= -1
formula for quadratic equations
k{x² -(sum of zeroes)x +(product of zeroes)}
k{x²-(-80/9)x+(-1)}
k{x²+80/9x-1}
k{9x²+80x-9}/9
let k=9
therefore, required equation is
9x²+80x-9
ANSWER:
- Required polynomial = 9x²+80x-9
GIVEN:
- First zero (α) = -9
- Second zero (β) = 1/9
TO FIND:
- A quadratic polynomial whose zero's are -9 and 1/9.
SOLUTION:
Standard form of Quadratic polynomial when roots are given:
= x²-(α+β)x+αβ ....(i)
Finding sum of zeros (α+β)
=> α = -9
=> β = 1/9
=> α+β = -9+1/9
=> α+β = (-81+1)/9
=> α+β = -80/9
Finding product of zeros (αβ)
=> αβ = -9*1/9
=> αβ = (-1)
Putting the values in eq (i)
= x²-(-80/9)x+(-1)
= x²+(80x/9) -1
Or:
=> x²+(80x/9)-1 = 0
=> (9x²+80x-9)/9 = 0
=> 9x²+80x-9 = 0
Required polynomial = 9x²+80x-9
NOTE:
Some important formulas:
=> Sum of zeros (α+β) = -(Coefficient of x)/Coefficient of x²
=> Product of zeros (αβ) = Constant term/ Coefficient of x²