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2x^2-px+q=0

x=1/2, -2

what is p-q?

 Jan 9, 2015

Best Answer 

 #2
avatar+26396 
+5

 x=1/2, -2      what is p-q? 

x1=12 and x2=2

 

2x2px+q=0|:2x2p2Px+q2Q=0x2+Px+Q=0\[[Vieta\]] :(x1+x2)=P=p2 and x1x2=Q=q2 pq=2P2Q=(2)[(x1+x2)]2x1x2=2(x1+x2x1x2)=2[12212(2)]=2(122+1)=1 pq=1

    

 Jan 9, 2015
 #1
avatar+130466 
+5

2x^2-px+q=0

x=1/2, -2

what is p-q?

We have that

2(1/2)^2 -(1/2)p + q = 0  →  (1/2) - (1/2)p + q = 0       (1) 

And .... 2(-2)^2 - (-2)p + q  = 0     →     8 + 2p  + q  = 0   (2)

And subtracting (1) from (2) we have  15/2 + (5/2)p = 0    →  15 + 5p =0  → p = -3

And using (2), we have 8 - 6 + q = 0   → 2 + q  = 0    →  = -2

So, our function is ..y =  2x^2 -3x - 2

So....p - q  = -3 - (-2)  = -1

 

 Jan 9, 2015
 #2
avatar+26396 
+5
Best Answer

 x=1/2, -2      what is p-q? 

x1=12 and x2=2

 

2x2px+q=0|:2x2p2Px+q2Q=0x2+Px+Q=0\[[Vieta\]] :(x1+x2)=P=p2 and x1x2=Q=q2 pq=2P2Q=(2)[(x1+x2)]2x1x2=2(x1+x2x1x2)=2[12212(2)]=2(122+1)=1 pq=1

    

heureka Jan 9, 2015

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