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what are 2 numbers that add to potitve 5 and times to give -7 

 Oct 16, 2021
 #1
avatar+391 
+1

Hello,

 

(\(x+y=5\)) and (\(xy=-7\)),

 

(\(x=5-y\)) and (\(xy=-7\)),

 

\((5-y)y=-7\)),

 

so y is now:

 

(\(y=\frac{5+\sqrt{53}}{2}\)) and (\(y=\frac{5-\sqrt{53}}{2}\)),

 

(\(x=5-\frac{5+\sqrt{53}}{2}\)) and (\(y=\frac{5-\sqrt{53}}{2}\)),

 

(\(x=5-\frac{5+\sqrt{53}}{2}\)) and (\(x=5-\frac{5-\sqrt{53}}{2}\)),

 

so x is now:

 

(\(x = \frac{5-\sqrt{53}}{2}\)) and (\(x = 5-\frac{5-\sqrt{53}}{2}\)),

 

(\(x = \frac{5-\sqrt{53}}{2}\)) and (\(x = \frac{5+\sqrt{53}}{2}\)),

 

So there are 2 possibilities, namely:

 

\(x,y=(\frac{5-\sqrt{53}}{2}, \frac{5+\sqrt{53}}{2})\) and the second possibility, namely:

 

\(x,y=(\frac{5+\sqrt{53}​​​​​​​​}{2}, \frac{5-\sqrt{53}}{2})\) 

 

Straight

 Oct 16, 2021
 #2
avatar+391 
+1

There is a hard methode to get \(\sqrt{53}\) , I have left it out

 Oct 16, 2021

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