Question

given that 1+2i is one root of equation x^4-3x^3+8x^2-7x+5=0 find other three roots

Answer 1

Answer:

these are the three roots of given

biquadratic eqution

Answer 2

Answer:

The roots of the given equation $x^4-3x^3+8x^2-7x+5=0.$ are 1+2i, 1-2i, 1+i√3, 1-i√3

Step-by-step explanation:

We are given that 1+2i is one root of equation $x^4-3x^3+8x^2-7x+5=0.$

We are asked to find the other three roots.

if 1+2i is the one root of given equation then 1-2i will be the another root.

With the help of these two roots, we can find the quadratic equation of the remaining roots and can find the remaining roots.

Sum of roots$=\frac{-b}{a}$

Here in this case $a=1, b=3, c=8, d=-7$ and $e=5$.

Substitute the values of a and b in the above formula.

Sum of roots  α+β+γ+δ$=\frac{-(-3)}{1}$

              1+2i+1-2i+γ+δ        $=3$

2+γ+δ=3

γ+δ=1

Product of roots = $=\frac{e}{a}$

Substitute the value of e and a.

α.β.γ.δ=5

(1-2i)(1+2i)(γδ)=5

γδ$=\frac{5}{5}$

Perform division.

γδ=1

Now, we know the sum of remaining roots and product of the remaining roots. So, we can write a quadratic equation.

x^2-(sum of roots)x+(product of roots)=0

$x^2-x(1)+1=0$

x=-b±√b^2-4ac/2a

x=-(-1)±√1-4(1)

x=1±i√3/2

The roots of the given equation are 1+2i, 1-2i, 1+i√3, 1-i√3