+0

# Algebra

+1
190
4

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

May 31, 2021

#2
+121095
+3

By Vieta

p + q  =   -5/1 =   - 5

pq  =  3/1   =   3

Note  that

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

May 31, 2021

#1
-1

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

May 31, 2021
#3
+309
0

WillBillDillPickle  Jun 1, 2021
edited by WillBillDillPickle  Jun 1, 2021
#2
+121095
+3

By Vieta

p + q  =   -5/1 =   - 5

pq  =  3/1   =   3

Note  that

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

CPhill May 31, 2021
#4
+26228
+2

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

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

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

Jun 1, 2021
edited by heureka  Jun 1, 2021