Question #8439c

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Answer 1 · 2016-03-23T01:44:12

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

$(2+sqrt34,2-sqrt34)$ , $(2-sqrt34,2+sqrt34)$