How do you solve the quadratic `x^2-(4+i)x+9+7i=0` using any method?

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Quadratic formula

The zeros of a quadratic in the form `ax^2+bx+c` are given by the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

In our example:

`x^2-(4+i)x+(9+7i)=0`

we have `a=1`, `b=-(4+i)`, `c=(9+7i)`, so:

`x = ((4+i)+-sqrt((-(4+i))^2-4(1)(9+7i)))/(2*1)`

`=((4+i)+-sqrt((16+8i+i^2)-36-28i))/2`

`=((4+i)+-sqrt(-21-20i))/2`

`=((4+i)+-i sqrt(21+20i))/2`

Sqaure root of a+bi

Note that I found in https://socratic.org/s/aw38evei

that the square roots of `a+bi` are:

`+-((sqrt((sqrt(a^2+b^2)+a)/2)) + (b/abs(b) sqrt((sqrt(a^2+b^2)-a)/2))i)`

In particular, if `a+bi` is in Q1, then its principal square root is

`(sqrt((sqrt(a^2+b^2)+a)/2)) + (sqrt((sqrt(a^2+b^2)-a)/2))i`

So we find:

`sqrt(21+20i)`

`=(sqrt((sqrt(21^2+20^2)+21)/2)) + (sqrt((sqrt(21^2+20^2)-21)/2))i`

`=(sqrt((sqrt(441+400)+21)/2)) + (sqrt((sqrt(441+400)-21)/2))i`

`=(sqrt((sqrt(841)+21)/2)) + (sqrt((sqrt(841)-21)/2))i`

`=(sqrt((29+21)/2)) + (sqrt((29-21)/2))i`

`=(sqrt(50/2)) + (sqrt(8/2))i`

`=(sqrt(25)) + (sqrt(4))i`

`=5+2i`

Conclusion

So:

`x=((4+i)+-i sqrt(21+20i))/2`

`=((4+i)+-i (5+2i))/2`

`=((4+i)+-(-2+5i))/2`

`= { ((2+6i)/2 = 1+3i), ((6-4i)/2 = 3-2i) :}`

`x^2-(4+i)x+(9+7i) = 0`

Note that if this has roots `x_1` and `x_2` then:

`x^2-(4+i)x+(9+7i)`

`= (x-x_1)(x-x_2) = x^2-(x_1+x_2)+x_1 x_2`

So we want to find `x_1`, `x_2` such that:

`{ (x_1 + x_2 = 4+i), (x_1 x_2 = 9+7i) :}`

Gaussian integers

`4+i` and `9+7i` are both Gaussian integers, that is Complex numbers of the form `a+bi` where `a` and `b` are integers.

What are the Gaussian integer factors of `9+7i`?

`|| 9 + 7i || = sqrt(9^2+7^2) = sqrt(81+49) = sqrt(130)`

Any Gaussian integer factors of `9+7i` must have norms with squares which are factors of `130`.

`130 = 2 * 5 * 13`

Up to a unit factor, the only Gaussian integer with norm `sqrt(2)` is `1+i`.

`(9+7i)/(1+i) = ((9+7i)(1-i))/((1+i)(1-i)) = (16-2i)/2 = 8-i`

Gaussian integers with norm `sqrt(5)` are `+-2+-i` and `+-1+-2i`. Up to a unit factor, we only need to try `2+i` and `2-i`:

`(8-i)/(2+i) = ((8-i)(2-i))/((2+i)(2-i)) = (15-10i)/5 = 3-2i`

So:

`9+7i = (1+i)(2+i)(3-2i)`

Any unit (`+-1` or `+-i`) multiple of these factors or their products is also a factor of `9+7i`.

Taking pairs of these factors we find:

`(1+i)(2+i) = 1+3i " "` with co-factor `(3-2i)`

`(2+i)(3-2i) = 8-i " "` with co-factor `(1+i)`

`(1+i)(3-2i) = 6-5i " "` with co-factor `(2+i)`

The first of these sums to `4+i` as required:

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

So our zeros are `1+3i` and `3-2i`