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Question 27789: two points (1,y)and (-1,x)lie on the curve y=px^2+qx+5 if y+x=14 what is the value of p?
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! two points (1,y)and (-1,x)lie on the curve y=px^2+qx+5 if y+x=14 what is the value of p?
The given curve is
y=px^2+qx+5 ----(1)
Given that y+x=14 ----(*) and two points (1,y) and (-1,x) lying on (1)
Required to find the value of p
A(1,y) is a point on (1) implies (putting x=1, y=y in (1) )
y=p1^2+qX1+5
y = pX1 +q +5
y = p+q+5 ----(2)
B(-1,x) is a point on (1) implies (putting x=-1, y=x in (1) )
that is putting x=-1, y=x in y=px^2+qx+5
x = pX(-1)^2 + qX(-1) +5
x =pX1-q+5
x =p-q +5 -----(3)
Adding (2) and (3) [why should we? because if we add, we get (y+x) and we can use the given data (x+y) = 14----(*) ]
(y+x) = (p+p) +(q-q) +(5+5)
(adding left with left and right with right term by term)
y+x = 2p +0 +10
14 = 2p +10 using (*)
14-10 = 2p {change side then change sign, taking 10 from the right to the left)
4 = 2p
2p =4
p = 4/2 = 2
Answer: p=2
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