Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the cofficient. 2√2xsquare–9x+5√2
Answers
Step-by-step explanation:
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Step-by-step explanation:
Given :-
2√2x²-9x+5√2
To find :-
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients ?
Solution :-
Finding zeroes :-
Given Quadratic Polynomial is 2√2x²-9x+5√2
Let P(x) = 2√2x²-9x+5√2
=> P(x) = 2√2x²-4x-5x+5√2
=> P(x) = 2√2x(x-√2)-5(x-√2)
=> P(x) = (x-√2)(2√2x-5)
To get zeroes we write P(x) = 0
=> (x-√2)(2√2x-5) = 0
=> x-√2 = 0 or 2√2x-5 = 0
=> x =√2 or 2√2 x = 5
=> x = √2 or x = 5/(2√2)
Zeroes are √2 and 5/(2√2)
Relationship between the zeroes and the coefficients:-
The zeroes are √2 and 5/(2√2)
Let α = √2 and β = 5/(2√2)
On Comparing P(x) with the standard quadratic Polynomial ax²+bx+c
We have
a = 2√2
b = -9
c = 5√2
i) Sum of the zeroes = α+β
=> α+β
= √2+5/(2√2)
= [(2√2×√2)+5]/(2√2)
= (4+5)/(2√2)
=> 9/(2√2)
= - ( Coefficient of x )/Coefficient of x²
= -b/a
α+ β = -b/a
ii) Product of the zeroes = αβ
=> (√2)(5/(2√2))
=5√2/(2√2)
= 5/2
= Constant term/ Coefficient of x²
= c/a
αβ = c/a
we get
Sum of the zeroes = -b/a
Product of the zeroes = c/a
Verified the relationship between the zeroes and the coefficients of P(x).
Answer:-
The zeroes are √2 and 5/(2√2)
Used formulae:-
- The standard quadratic Polynomial ax²+bx+c
- Sum of the zeroes = -b/a
- Product of the zeroes = c/a