An equation with real coefficients has 1-i, 3 and 4 among it's roots. What is the lowest possible degree of the equations?
if (1-i) then also (1+i)
x^2 --> 2 roots
x^3 ---> 3 roots
we have 4 roots so x^4
To find the lowest possible degree of the equation, we need to consider the complex conjugate of 1-i, which is 1+i.
Since complex roots always come in conjugate pairs, the equation must have the roots 1-i and 1+i.
The equation must also have the roots 3 and 4.
Therefore, the equation has at least four distinct roots: 1-i, 1+i, 3, and 4.
To have four distinct roots, the equation must be of degree four.
Hence, the lowest possible degree of the equation is four.
To find the lowest possible degree of the equation, we need to consider the properties of complex roots.
If an equation has complex roots, they always come in conjugate pairs. This means that if one complex number, a + bi, is a root of the equation, its conjugate a - bi is also a root.
In this case, we have the complex root 1 - i. Therefore, its conjugate, 1 + i, must also be a root of the equation.
Now, let's consider the real roots. We have the real roots 3 and 4.
To form an equation with these given roots, we can start by factoring out the linear factors for each of the roots:
For the complex roots, we have:
(x - (1 - i))(x - (1 + i)) = ((x - 1) + i)((x - 1) - i)
= ((x - 1)^2 - i^2)
= (x - 1)^2 + 1
For the real roots, we have:
(x - 3)(x - 4) = x^2 - 7x + 12
Multiplying these factors, we get:
((x - 1)^2 + 1)(x^2 - 7x + 12)
Expanding this expression, we get a polynomial equation that satisfies the given roots. Simplifying further, we get:
x^4 - 9x^3 + 26x^2 - 24x + 65
So, the lowest possible degree of the equation is 4.