Question #8439c

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Question #8439c

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Answer 1 · 2016-03-23T01:44:12

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

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

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

we get

$(4 - b) \cdot b = - 30 \Rightarrow 4 b - b^{2} = - 30 \Rightarrow b^{2} - 4 b - 30 = 0 \Rightarrow {(b - 2)}^{2} - 34 = 0 \Rightarrow b_{1} = 2 + \sqrt{34} or b_{2} = 2 - \sqrt{34}$ $(4-b)*b=-30=>4b-b^2=-30=>b^2-4b-30=0=> (b-2)^2-34=0=>b_1=2+sqrt34 or b_2=2-sqrt34$

Hence $a_{1} = 2 - \sqrt{34}$ $a_1=2-sqrt34$ and $a_{2} = 2 + \sqrt{34}$ $a_2=2+sqrt34$

Finally we have the following pairs of solutions

$(2 + \sqrt{34}, 2 - \sqrt{34})$ $(2+sqrt34,2-sqrt34)$ , $(2 - \sqrt{34}, 2 + \sqrt{34})$ $(2-sqrt34,2+sqrt34)$

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Question #8439c

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