Find the zeroes of the polynomial.x2-3 and verify the relationship between the zeroes and the coefficients.
Answers
Step-by-step explanation:
Given :-
The polynomial is X²-3
To find :-
Find the zeroes of the polynomial x²-3 and verify the relationship between the zeroes and the coefficients.?
Solution :-
Given that
The quardratic polynomial P(x) = X²-3
Finding the zeroes :-
To get zeroes of P(x) then we equate P(x) to zero
=> P(x) = 0
=> X²-3 = 0
=> X²-(√3)² = 0
=> (X+√3)(X-√3) = 0
Since, (a+b)(a-b) = a²-b²
Where, a = X and b = √3
=> X+√3 = 0 (or) X-√3 = 0
=> X = -√3 (or) X = √3
The zeroes are √3 and -√3
Verifying the relationship between the zeroes and the coefficients :-
Given quardratic polynomial P(x) = X²-3
On comparing with the standard quadratic polynomial ax²+bx+c then
a = 1
b = 0
c = -3
Zeroes are √3 and -√3
Sum of the zeroes = √3+(-√3)
=> (√3-√3)
=> 0
=> -(Coefficient of x)/Coefficient of x²
=> -(0)/1
=> 0/1
=> 0
=> -b/a
Sum of the zeroes = -b/a
Product of the zeroes = (√3)(-√3)
=> -[√(3×3)]
=> -√9
=> -3
=> Constant term / Coefficient of x²
=> c/a
Product of the zeroes = c/a
Verified.
Used formulae:-
→ The standard quadratic Polynomial is ax²+bx+c
→ Sum of the zeroes = -b/a
→ Product of the zeroes = c/a
→ (a+b)(a-b) = a²-b²