If (3+√2)/2 and (3-√2)/2 are the roots of the polynomial 2x² + ax + b. Find the value of a and b.
Answers
Step-by-step explanation:
Given :-
(3+√2)/2 and (3-√2)/2 are the roots of the polynomial 2x² + ax + b.
To find :-
Find the value of a and b ?
Solution :-
Method-1:-
Given Quadratic Polynomial P(x) = 2x²+ax+b
On Comparing this with the standard quadratic Polynomial ax²+bx+c
a = 2
b = a
c = b
Given zeroes are (3+√2)/2 and (3-√2)/2
We know that
Sum of the zeroes = -b/a
=> [(3+√2)/2]+[(3-√2)/2] = -a/2
=> (3+√2+3-√2)/2 = -a/2
=> (3+3)/2 = -a/2
=> 6/2 = -a/2
=> 6 = -a
=> a = -6
And
Product of the zeroes = c/a
=> [(3+√2)/2][(3-√2)/2] = b/2
=> [(3+√2)(3-√2]/(2×2) = b/2
=>[ {(3)²-(√2)²}/2²] = b/2
Since (a+b)(a-b) = a²-b²
=> (9-2)/4 = b/2
=> 7/4 = b/2
=> b/2 = 7/4
=> b = (7/4)×2
=> b = 7/2
Therefore, a = -6 and b = 7/2
Method -2:-
Given Quadratic Polynomial P(x) = 2x²+ax+b
Given zeroes = (3+√2)/2 and (3-√2)/2
Let α = (3+√2)/2 and β = (3-√2)/2
We know that
The Quadratic Polynomial whose zeores are α and β is K[x²-(α+β)x+α β]
=> K[x²-[(3+√2)/2 +(3-√2)/2]x+[(3+√2)/2][(3-√2)/2]]
=>K[x²-[(3+3)/2]x+{(3²-(√2)²}/4]
=> K[x²-(6/2)x+(9-2)/4]
=> K[x²-3x+(7/4)]
=> K[(4x²-12x+7)/4]
If K = 4 then the quardratic polynomial 4x²-12x+7
Given polynomial = 2x²+ax+b
=> 2x²+ax+b = 4x²-12x+7
=> 2x²+ax+b =2x²-(12/2)x+(7/2)
=> 2x²+ax+b =2x²-6x+(7/2)
On Comparing both sides then
=> a = -6 and b = 7/2
Therefore, a = -6 and b = 7/2
Answer:-
The value of a = -6
The value of b = 7/2 for the given problem
Used formulae:-
- The standard quadratic polynomial is ax²+bx+c
- Sum of the zeroes = -b/a
- Product of the zeroes = c/a
- The Quadratic Polynomial whose zeores are α and β is K[x²-(α+β)x+α β]
- (a+b)(a-b) = a²-b²