How do you solve the system #x^2 + y^2 = 13#, # y=x^2 - 1#?

Question

6628 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-18T00:07:26

Use the second equation to substitute for #x^2# in the first equation to get a quadratic in #y#. Solve for #y# and hence for #x# to find solutions:

#(+-2, 3)# and #(+-sqrt(3)i, -4)#

Add #1# to both sides of the second equation to get:

#x^2 = y+1#

Substitute this expression for #x^2# into the first equation to get:

#(y+1)+y^2=13#

Subtract #13# from both sides and rearrange to get:

#y^2+y-12 = 0#

Note that #4xx3 = 12# and #4-3 = 1#

Hence:

#0 = y^2+y-12 = (y+4)(y-3)#

So #y = 3# or #y = -4#

So #x^2 = y+1 = 4# or #x^2 = y+1 = -3#

So #x = +-sqrt(4) = +-2# or #x = +-sqrt(-3) = +-sqrt(3)i#

So the possible Real solutions are #(+-2, 3)# and the possible Complex solutions are #(+-sqrt(3)i, -4)#