Question

If the biquadratic x⁴+ax³ +bx² +cx+d=0(a,b,c,dϵR) has 4 non real roots, two with sum 3+4i and the other two with product 13+i. Find the sum of the digits of value of 'b'​

Answer 1

Answer:

x4 + ax3 + bx2 + cx + d has 4 non-real roots

Step-by-step explanation:

Let the roots of the equation (1) be α±iβ and γ±iδ

So, the equation will be

(x−(α+iβ))(x−(α−iβ))(x−(γ+iδ))(x−(γ−iδ))=0

(x2−2αx+α2+β2)(x2−2γx+γ2+δ2)=0

Here, we have to find b

So, the coefficient of x2 terms in above equation will give b.

b=α2+β2+4αγ+γ2+δ2    ..(2

Now, given α+iβ+γ+iδ=3+4i

On comparing, we get

α+γ=3 and β+δ=4

On squaring, we get

α2+γ2+2αγ=9       ....(3)

β2+δ2+2βδ=16

⇒β2+δ2=16−2βδ       

Using (3) and (4) in (2), we get

b=9+2αγ+16−2βδ     ....(5)

Also, given (α−iβ)(γ−iδ)=13+i

⇒(αγ−βδ)−i(αδ+βγ)=13+i

On comparing, we get

αγ−βδ=13

Put this value in (5),we get

b=25+2(13)

b=51

Sum of digits is 6.