quadratic expression and equations .....................if u dont know then plz dont make spams
HINT!ANSWER IS 25
Answers
Given :-
- ɑ and β are Roots of The Equation x² - x + 2 = 0
To Find :-
- [ɑ^(-6) + β^(-6) + 2ɑ(-3) * β^(-3) ] ɑ^6 * β^6
Solution :-
The sum of the roots of the Equation ax² + bx + c = 0 , is given by = (-b/a)
and ,
→ Product of roots of the Equation is given by = c/a.
So, Comparing x² - x + 2 = 0 , we get,
→ a = 1
→ b = (-1)
→ c = 2
So,
→ sum of roots = (ɑ + β) = (-b/a) = -(-1)/1 = 1
→ Product of roots = ɑ * β = (c/a) = (2/1) = 2
_____________________
Now,
→ [ɑ^(-6) + β^(-6) + 2ɑ(-3) * β^(-3) ] ɑ^6 * β^6
→ { ɑ^(-6) * ɑ^6 * β^6 } + { β^(-6) * ɑ^6 * β^6 } + { 2ɑ(-3) * β^(-3) * ɑ^6 * β^6 }
→ { β^6 + ɑ^6 + 2ɑ³ * β³ }
Comparing it with (a² + b² + 2ab ) = (a +b)²
→ ( ɑ³ + β³)²
→ [ (ɑ + β)³ - 3ɑ * β(ɑ + β) ]² { (a³ + b³ = (a + b)³ - 3ab(a+b) }
Putting values from both Sum and product now, we get,
→ [ ( 1)³ - 3 * 2(1) ]²
→ [ 1 - 6]²
→ (-5)²
→ 25 (Ans).
[ Easiest Question ].
Note;I am taking alpha=a and beta =b
now, let's move on questiion
a+b=1......i) (hope u remembered the formula of sum and product of roots)
ab=2
now,
simplify the given value to be taken out
on simplifying we get
b^6+a^6+2a^3b^3
which is in the form
(b^3)^2 +(a^3)^2+2a^3b^3
(a^3+b^3)^2......ii)
from first we get
(a+b)^3=1
a^3+b^3+3ab(a+b)=1
a^3+b^3=1-6=-5
putting this value in ii) we get
(-5)^2
=25
hope it helps