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?
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.