Question

1,2,3,4 are the zeroes of x^4+ax³+6x²+cx+d=0 then the value of a is ..............(^ is denoted for power of ) *​

Answer 1

Step-by-step explanation:

Let's solve for a.

x4+ax3+6x2+cx+d=0

Step 1: Add -x^4 to both sides.

ax3+x4+cx+6x2+d+−x4=0+−x4

ax3+cx+6x2+d=−x4

Step 2: Add -cx to both sides.

ax3+cx+6x2+d+−cx=−x4+−cx

ax3+6x2+d=−x4−cx

Step 3: Add -6x^2 to both sides.

ax3+6x2+d+−6x2=−x4−cx+−6x2

ax3+d=−x4−cx−6x2

Step 4: Add -d to both sides.

ax3+d+−d=−x4−cx−6x2+−d

ax3=−x4−cx−6x2−d

Step 5: Divide both sides by x^3.

ax3/x3=−x4−cx−6x2−d/x3

a=−x4−cx−6x2−d/x3

Answer 2

Step-by-step explanation:

ANSWER

Given equation is

x

4

+ax

3

+bx

2

+cx+d=0 .....(1)

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

⇒(x

2

−2αx+α

2

+β

2

)(x

2

−2γx+γ

2

+δ

2

)=0

Here, we have to find b

So, the coefficient of x

2

terms in above equation will give b.

⇒b=α

2

+β

2

+4αγ+γ

2

+δ

2

....(2)

Now, given α+iβ+γ+iδ=3+4i

On comparing, we get

α+γ=3 and β+δ=4

On squaring, we get

α

2

+γ

2

+2αγ=9 ....(3)

β

2

+δ

2

+2βδ=16

⇒β

2

+δ

2

=16−2βδ .....(4)

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.