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The roots of x^2 + 5x + 3 = 0 are p and q.  Find 1/p + 1/q.

 May 31, 2021

Best Answer 

 #2
avatar+121056 
+3

By Vieta

 

p + q  =   -5/1 =   - 5

 

pq  =  3/1   =   3

 

Note  that

 

1/p  +  1/q  =     (p + q)  / (pq)  =       -5 / 3

 

 

cool cool cool

 May 31, 2021
 #1
avatar
-1

1/p + 1/q = 3/14.

 May 31, 2021
 #3
avatar+309 
0

Your wrongfrownblushangrysadsurprise

 

cheeky

WillBillDillPickle  Jun 1, 2021
edited by WillBillDillPickle  Jun 1, 2021
 #2
avatar+121056 
+3
Best Answer

By Vieta

 

p + q  =   -5/1 =   - 5

 

pq  =  3/1   =   3

 

Note  that

 

1/p  +  1/q  =     (p + q)  / (pq)  =       -5 / 3

 

 

cool cool cool

CPhill May 31, 2021
 #4
avatar+26213 
+2

The roots of \(x^2 + 5x + 3 = 0\) are p and q.
Find \(\dfrac1p + \dfrac1q\).

 

\(\text{Set $x=\dfrac{1}{x}$ in $x^2 + 5x + 3$ }\)

 

\(\begin{array}{|rcll|} \hline x^2 + 5x + 3 &=& 0 \\\\ \Rightarrow \left(\dfrac{1}{x}\right)^2 + 5*\dfrac{1}{x} + 3 &=& 0 \\\\ \dfrac{1}{x^2} + \dfrac{5}{x} + 3 &=& 0 \quad | \quad * x^2 \\\\ 1 + 5x + 3x^2 &=& 0 \\\\ 3x^2+5x+1 &=& 0 \quad | \quad : 3 \\\\ x^2+\dfrac{5}{3}x +\dfrac13 &=& 0 \\ \hline \text{By Vieta:}~\dfrac1p + \dfrac1q &=& -\dfrac{5}{3} \\\\ \text{By Vieta:}~\dfrac1p * \dfrac1q &=& \dfrac{1}{3} \\ \hline \end{array}\)

 

laugh

 Jun 1, 2021
edited by heureka  Jun 1, 2021

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