Math, asked by Anonymous, 10 months ago

quadratic expression and equations .....................if u dont know then plz dont make spams
HINT!ANSWER IS 25​

Attachments:

Answers

Answered by RvChaudharY50
137

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 - 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 ( + + 2ab ) = (a +b)²

( ɑ³ + β³)²

→ [ (ɑ + β)³ - 3ɑ * β(ɑ + β) ]² { ( + = (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 ].

Answered by Rajshuklakld
147

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

Similar questions