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Question 224409:
if 2x^2 - 14x + 29 = p(x+q)^2 +r holds for all real values of x, find the values of p,q and r. Hence, state the minimum value of the function and the corresponding value of x..
im stuck.. plz help :(
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! if 2x^2 - 14x + 29 = p(x+q)^2 +r holds for all real values of x, find the values of p,q and r. Hence, state the minimum value of the function and the corresponding value of x..
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2x^2 - 14x + 29 = px^2 + 2pqx +pq^2 +r
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Matching up the coefficients you get:
p = 2
2pq= -14
pq^2+r = 29
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So:
p = 2
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q = (14/4) = 7/2
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2(7/2)^2+r = 29
(49/2) + r = (58/2)
r = 9/2
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Hence, state the minimum value of the function and the corresponding value of x
a = 2 ; b = -14
min occurs at x = -b/2a = 14/4 = 7/2
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2x^2 - 14x + 29
Minimum value is 2(29)^2 - 14*29 +29 = 1305
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Cheers,
Stan H.
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