Question

Let a and B are the roots of the equation
x^2 + x + 1 = 0 then
(1) a² + B² = 4
(2) (a – B)^2 = 3
(3) a^3 + b^3 = 2
(4) a^4 +b^4 = 1
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Answer 1

Answer:

Note:

Let's us consider a quadratic equation in variable x , say: ax^2 + bx + c = 0

If A and B are its roots, then ;

A + B = - b/a

A•B = c/a

Here, the given quadratic equation is ;

x^2 + x +1 = 0

Clearly, we have;

a = 1

b = 1

c = 1

It is given that , A and B are the roots of the given quadratic equation.

Thus;

A + B = - b/a = -1/1 = -1

A•B = c/a = 1/1 = 1

Now , let's find the following quantities one-by-one ;(A^2 + B^2) , (A - B)^2 ,

(A^3 + B^3) , (A^4 + B^4)

1) we know that,

=> (A + B)^2 = A^2 + B^2 + 2•A•B

=> (-1)^2 = A^2 + B^2 + 2•1

=> A^2 + B^2 = 1 - 2

=> A^2 + B^2 = -1

{ this is not possible , as the sum of two positive numbers is never negative}

2) we know that,

=> (A - B)^2 = A^2 + B^2 - 2•A•B

=> (A - B)^2 = -1 - 2•1

=> (A - B)^2 = -3

{ this is not possible, as the square of any number is never negative}

3) we know that,

=> A^3 + B^3 = (A+B)(A^2 + B^2 - A•B)

=> A^3 + B^3 = -1( -1 -1)

=> A^3 + B^3 = 2

4) we know that,

=> (A^2+B^2)^2=A^4 + B^4 + 2(A^2)(B^2)

=> (-1)^2 = A^4 + B^4 + 2•1

=> A^4 + B^4 = 1 - 2

=> A^4 + B^4 = -1

{ this is not possible, as the sum of positive numbers is never negative}

Thus , the answer is ;

option:(3)

Moreover,

The discriminant of this equation is ;

D = b^2 - 4ac

D = 1^2 - 4•1•1

D = -3

D < 0

Here, the discriminant is less then zero, thus both of its roots are imaginary.