7 If p(x) = (ax 2 + bx +c) has zeroes as ‘α’ and ‘β’, find the values of each of the following:- (a) 1 α + 1 β (b) α 3 + β 3 (c) 1 α 3 + 1 β 3 (d) α - β (e) α 3 - β 3
Answers
Step-by-step explanation:
Given :-
P(x) = ax²+bx+c has zeroes are ‘α’ and ‘β’
To find :-
Find the values of each of the following
a)1/α + 1/β
(b) α³+ β³
(c) 1 /α³+ 1 / β³
(d) α - β
(e) α³ - β³
Solution :-
Given that
The Quadratic Polynomial P(x) = ax²+bx+c
Given zeroes are ‘α’ and ‘β’
We know that
Sum of the zeroes = -b/a
α + β = -b/a --------(1)
Product of the zeroes = c/a
=> αβ = c/a ---------(2)
a)Finding the value of 1/α + 1/β :-
1/α + 1/β
=> (α + β)/αβ
=> (-b/a)/(c/a)
=> (-b/a)×(a/c)
=> -ab/ac
=> -b/c
b)Finding the value of α³+ β³:-
We know that
a³+b³ = (a+b)³-3ab(a+b)
α³+ β³ = (α+ β)³ -3αβ(α+ β)
=> α³+ β³ = (-b/a)³-3(c/a)(-b/a)
=>α³+ β³ = (-b³/a³)+(3bc/a²)
=> α³+ β³ = (-b³+3abc)/a³
c)Finding the value of 1 /α³+ 1 / β³
1 /α³+ 1 / β³
=>( α³+ β³)/(α³β³)
=> [(-b³+3abc)/a³]/(c/a)³
=> [ (-b³+3abc)/a³]×(a³/c³)
=> (-b³+3abc)×c³ or
=> -b³c³+3abc⁴
d)Finding the value of α - β:-
We know that
(a-b)² = (a+b)²-4ab
(α - β)² = (α + β)²-4αβ
=> (α - β)² = (-b/a)²-4(c/a)
=> (α - β)² = (b²/a²)-(4c/a)
=> (α - β)² = (b²-4ac)/a²
=> α- β=√[(b²-4ac)/a²]
=>α- β = [√(b²-4ac)] /a
and
(a+b)² = a²+2ab+b²
(α + β)² = α²+ β²+2α β
=> (-b/a)² = α²+ β²+2(c/a)
=> b²/a² = α²+ β²+(2c/a)
=> α²+ β² = (b²/a²)-(2c/a)
=> α²+ β² = (b²-2ac)/a²
e) Finding the value of α³ - β³:-
We know that
a³-b³ = (a-b)(a²+ab+b²)
α³- β³ = (α - β)(α²+ β²+α β)
=> α³- β³ = [√(b²-4ac)/a][((b²-2ac)/a²)+(c/a)]
=> α³- β³ = [√(b²-4ac)/a][(b²-2ac+ac)/a²]
=>α³- β³ = [√(b²-4ac)/a][(b²-ac)/a²]
=> α³- β³ = [√(b²-4ac)](b²-ac)/a³
Answer :-
a) 1/α + 1/β = -b/c
b)α³+ β³ = (-b³+3abc)/a³
c)1 /α³+ 1 / β³ = (-b³+3abc)×c³ or -b³c³+3abc⁴
d)α- β = [√(b²-4ac)] /a
e)α³- β³ = [√(b²-4ac)](b²-ac)/a³
Used formulae:-
- (a+b)² = a²+2ab+b²
- a³-b³ = (a-b)(a²+ab+b²)
- (a-b)² = (a+b)²-4ab
- a³+b³ = (a+b)³-3ab(a+b)
- The Quadratic Polynomial P(x) = ax²+bx+c
- Sum of the zeroes = -b/a
- Product of the zeroes = c/a