Question

Question #8439c

Answer 1

Let `a`,`b` is the real two numbers hence

`a+b=4` and `a*b=-30`

From the first one we have that `a=4-b` substitute to the second

we get

#(4-b)*b=-30=>4b-b^2=-30=>b^2-4b-30=0=> (b-2)^2-34=0=>b_1=2+sqrt34 or b_2=2-sqrt34#

Hence `a_1=2-sqrt34` and `a_2=2+sqrt34`

Finally we have the following pairs of solutions

`(2+sqrt34,2-sqrt34)` ,`(2-sqrt34,2+sqrt34)`