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#