# Question 8439c

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

$a + b = 4$ and $a \cdot b = - 30$

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 ${a}_{1} = 2 - \sqrt{34}$ and ${a}_{2} = 2 + \sqrt{34}$

Finally we have the following pairs of solutions

$\left(2 + \sqrt{34} , 2 - \sqrt{34}\right)$ ,$\left(2 - \sqrt{34} , 2 + \sqrt{34}\right)$