# If 6 + i and a^2 - a + b*i are the roots of the same quadratic equation, what is the value of a and b?

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### 3 Answers

If a quadratic equation ax^2 + bx + c = 0, has one complex root a + bi, the other root has to be equal to a – bi.

Here one of the roots is 6 + i, so the other root is 6 – i.

Now 6 – i = a^2 - a + b*i

Equating the real coefficient

a^2 – a = 6

=> a^2 – a – 6 = 0

=> a^2 – 3a + 2a – 6 =0

=> a (a – 3) + 2(a – 3) = 0

=> (a + 2) (a – 3) = 0

So a can be -2 and 3.

Also, equating the imaginary coefficient b = -1

**Therefore a is equal to -2 and 3 and b is -1.**

If a root of a quadratic is complex, then the other root is also a complex root and it is the conjugate of the first root.

If z1 = 6 + i, then z2 = 6 - i

But, from enunciation, we know that a^2 - a + b*i = 6 - i.

The real parts from both sides are equal, also the imaginary parts are equal.

a^2 - a = 6

We'll subtract 6 both sides:

a^2 - a - 6 = 0

We'll apply quadratic formula:

a1 = [1 + sqrt(1 + 24)]/2

a1 = (1 + 5)/2

**a1 = 3**

**a2 = -2**

**b = -1**

We know that the complex roots of a polynomial always occrs in pairs and the pairs are the conjugates of each other.

So if x+yi is a complex root of a polynomial, then the conjugate of x+yi is x-yi. Then x-yi is also a root of the polynomial.

In the given, 6+i is a complex root of a quadratic. So 6-i is also a complex root of the quadratic equation. So a^2-b+bi = 6-i.

a^2-2 +bi = 6-i.

We equate the real parts on both sides and similarly equate the imaginary parts on both sides:

=> a^2-a = 6 and b = -1.

a^2-a = 6 => (a-3)(a+2) = 0.

So a = 3, or a = -2.

Therefore a = 3, or a= -2, b = -1.