# Determine a and b and solve the equation x^4-7x^3+21x^2+ax+b=0 if 1+2i is the root of equation.

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### 3 Answers

x^4-7x^3+21x^2+ax+b = 0 has a root 1+2i .

Therefore the conjugate of 1+2i is also a root of the given equation.

The conjugate of 1+2i is 1-2i.

Therefore , (x-(1+2i)) and (x-(1-2i)) are the factors of x^4-7x^3+21x^2+ax+b by remainder theorem.

(x-1-2i)(x-1+2i) = (x-1)^2 - (2i)^2 = (x^2-2x+1+4) = (x^2+5x+5) is also a factor.

Therefore the x^2-7x^3+21x^2 +ax+b should be divisible by (x^2-2x+5).

So we divide x^4-7x^3+21x^2+ax+b by x^2-2x+5.

x^2-2x+5)x^4-7x^3+21x+ax+b( x^2

x^4 -2x^3 +5x^2

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x^2-2x+5) -5x^3+16x^2 +ax+b ( x^2-5x

-5x^3 +10x^2 -25x

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x^2-2x+5)6x^2+(a+25)x+b (x^2-5x+6

6x^2 -12x +30

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( a+25 +12)x +(b-30)

So the remainder = (a+37)x +(b-30) .

Since x^2-2x+5 is factor of x^4-7x^3+21x^2+ax+b, the remainder (a+37)x +(b-30) should be zero.

Therefore the coefficient of x , a+37= 0 and the contant term b-30 = 0.

Therefore a = -37 and b = 30.

So the given equation should be x^2-7x^3+21x^2-37x+30

We are given the equation x^4 - 7x^3 + 21x^2 + ax + b = 0 and we have to find a and b if one of the roots of the equation is 1+ 2i.

Now as 1+2i is a complex root 1-2i is also a root.

(1+2i)^2 = 1 + 4i^2 + 4i = 1- 4 + 4i = -3 + 4i

(1-2i)^2 = 1 + 4i^2 - 4i = 1- 4 - 4i = -3 - 4i

(1+2i)^3 = (-3+4i)(1+2i)= -3-6i+4i-8 = -11 -2i

(1-2i)^3 = (-3-4i)(1-2i)= -3+6i-4i-8 = -11 +2i

(1+2i)^4 = (-3+4i)^2 = -7 – 24i

(1+2i)^4 = (-3-4i)^2 = -7 + 24i

Now substituting the root x= 1+ 2i in the equation we have :

-7 – 24i – 7*(-11 -2i)+ 21(-3 + 4i) + a(1+2i) +b =0

=> -7 – 24i + 77 + 14i – 63 + 84i+ a + 2ai +b =0

=>7 +a + b + 74i + 2ai =0

=> 2a + 74 = 0

=> a = -37

=> and as 7+ a + b = 30

=> b = 30

Therefore a = -37, b= 30

**So the equation is x^4 - 7x^3 + 21x^2 + -37x + 30 = 0**

If 1+2i is the root of equation then, according to the rule, the conjugate of 1+2i is the root of equation, also.

The conjugate of 1+2i is 1 - 2i.

A value represents the root of an equation, if and only if, substituted in the equation, it cancels the equation.

7x^3+21x^2+ax+b=0

We'll substitute x by 1+2i:

7(1+2i)^3+21(1+2i)^2+a(1+2i)+b=0

We'll factorize by 7(1+2i)^2 the first 2 terms:

7(1+2i)^2*(1 + 2i + 3) + a(1+2i) + b = 0

We'll expand the square and we'll combine like terms:

7(1 + 4i - 4)(4 + 2i) + a(1+2i) + b = 0

7(-3 + 4i)(4 + 2i) + a(1+2i) + b = 0

We'll remove the brackets:

-84 - 42i + 112i - 56 + a + 2ai + b = 0

We'll combine the real parts and the imaginary parts:

-140 + b + a + i(70 + 2a) = 0

Th real parts from both sides have to be equal:

-140 + b + a = 0

a + b = 140 (1)

The imaginary parts from both sides have to be equal:

70 + 2a = 0

2a = -70

**a = -35 **

We'll substitute a in (1):

a + b = 140

-35 + b = 140

b = 140 + 35

**b = 175**

**The equation, whose root is 1 + 2i, is:**

**7x^3 + 21x^2 - 35x + 175 = 0**