if a and ß are zeroes of quadratic polynomial 2x²+5x+2 then find the value of a⁴+ß⁴
Answers
Step-by-step explanation:
Given :-
a and ß are zeroes of quadratic polynomial 2x²+5x+2 .
To find :-
Find the value of a⁴+ß⁴?
Solution:-
Given Quadratic Polynomial=P(x) = 2x²+5x+2
On Comparing this with the standard quadratic Polynomial ax²+bx+c
We have,
a = 2
b = 5
c = 2
Given zeroes = a and ß
We know that
Sum of the zeroes = -b/a
=> a+ß = -5/2 -------(1)
=> ß = (-5/2)-a ------(2)
and
Product of the zeroes = c/a
=> a ß = 2/2
=> aß = 1-------(3)
We know that
(a-b)² = (a+b)²-4ab
=> (a-ß)² = (a+ß)²-4(aß)
=> (a-ß)²= (-5/2)²-4(1)
=>(a-ß)²= (25/4)-4
=>(a-ß)²=(25-16)/4
=> (a-ß)² = 9/4
=>(a-ß) =3/2---(4) (taking positive value )
On adding (1)&(4) then
a+ß+a-ß = (-5/2)+(3/2)
=> 2a = (-5+3)/2
=> 2a = -2/2
=> 2a = -1
=> a = -1/2
On Substituting the value of a in (2) then
ß = (-5/2)-(-1/2)
=> ß = (-5/2)+(1/2)
=> ß = (-5+1)/2
=> ß = -4/2
=> ß = -2
We have , a =-1/2 ,ß = -2
The value of a⁴+ß⁴
=> (-1/2)⁴+(2)⁴
=> (1/16)+16
=>(1+16×16)/16
=>(1+256)/16
=>257/16
Answer:-
The value of a⁴+ß⁴ for the given problem is 257/16
Used formulae:-
- The standard quadratic polynomial is ax²+bx+c
- Sum of the zeroes = -b/a
- Product of the zeroes = c/a
- (a-b)² = (a+b)²-4ab