if a and b are the roots of 3x^2-6x+2=0, then find a) a^2+b^2 b) a^3+b^3
Answers
Given :-
a and b are the roots of 3x²-6x+2=0
To find :-
The values of a) a²+b² b) a³+b³
Solution :-
Given quadratic equation is 3x²-6x+2 = 0
On comparing this with the standard quadratic equation ax²+bx+c = 0
a = 3
b = -6
c = 2
Given roots are a and b
We know that
Sum of the roots = -b/a
Therefore, a+b = -(-6/3)
=> a+b = 6/3
=> a+b = 2 -------(1)
Product of the roots = c/a
=> ab = 2/3 -------(2)
Now,
The value of a²+b²
=> (a+b)²-2ab
=> 2²-2(2/3)
=> 4-(4/3)
=> (12-4)/3
=> 8/3
and
The value of a³+b³
=> (a+b)(a²+b²-ab)
=> (2)[(8/3)-(2/3)]
=> (2)(8-2)/3
=> (2)(6/3)
=> (2)(2)
=> 4
or
a³+b³ = (a+b)³-3ab(a+b)
=> a³+b³ = (2)³-3(2/3)(2)
=> a³+b³ = 8-4
=> a³+b³ = 4
Answer :-
1) The value of a²+b² is 8/3
2) The value of a³+b³ is 4
Used formulae:-
→ The standard quadratic equation is
ax²+bx+c = 0
→ Sum of the roots = -b/a
→ Product of the roots = c/a
→ a²+b² = (a+b)²-2ab
→ a³+b³ = (a+b)(a²-ab+b²) = (a+b)³-3ab(a+b)
refer the given attachment
