Question

-1+i is a root of x^4+4x^3+5x^2+2x+k=0 then the other roots are?​

Answer 1

Step-by-step explanation:

Given:

Let x2 + 2 = t

⇒ t2 + 8x2 = 6xt

⇒ t2 - 6xt + 8x2 = 0

⇒ t2 - 4xt - 2xt + 8x2 = 0

⇒ t (t – 4x) – 2x (t – 4x) = 0

⇒ (t – 2x) (t – 4x) = 0

Now, putting the value of t in above equation, we get

⇒ (x2 + 2 – 2x) (x2 + 2 – 4x) = 0

Now, (x2 – 2x + 2) = 0 and (x2 – 4x + 2) = 0

⇒ x2 – 2x + 2 = 0

Roots are imaginary.

Now,

⇒ (x2 – 4x + 2) = 0

Roots are real.

We can say that not all roots of the equation are complex.

So statement 1st is wrong.

Now,

Sum of all the roots = (1 + i) + (1 – i) + (2 + √2) + (2 - √2) = 6

So statement 2nd is true.