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$0 = 9a + 3b + c$

$2 = 25a + 5b + c$

$6 = 49a + 7b + c$

These are my 3 equations that have 3 unknowns. How do I solve for these unknowns?

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What does education/quadtric-forms/problem-solving/quadratics have anything to do with this? –  Inquest Oct 18 '12 at 23:53
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@Inquest See my answer. –  Bill Dubuque Oct 19 '12 at 3:08
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2 Answers 2

up vote 3 down vote accepted

(1) $9a+3b+c=0$

(2) $25a+5b+c=2$

(3) $49a+7b+c=6$

Using (2)-(1) we have:

(4) $16a+2b=2$

Using (3)-(2) we have:

(5) $24a+2b=4$

Using (5)-(4) we have:

(6) $8a=2$

It follows that $a=\frac{1}{4}$, $b=-1$ and $c=\frac{3}{4}$.

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Since you had tagged your question "quadratic" and "quadratic-forms" (removed by another user) I will interpret that as a hint provided to you to exploit the quadratic structure of the problem. Notice that the problem can be viewed as interpolating a quadratic through $3$ points, as follows.

Let $\rm\:f(x) = a\,x^2 + b\,x + c$

$\rm 0 = f(3)\:\Rightarrow\: f(x) = (x-3)(ax-d),\ $ where $\rm\,\ 3d = c = f(0)$
$\rm 2 = f(5)\, =\, 2(5a-d)\:\Rightarrow\:5a-d = 1 $
$\rm 6 = f(7)\, =\, 4(7a-d)\:\Rightarrow\:7a-d = 3/2\:\Rightarrow\:2a = \frac{1}2\:\Rightarrow\:a=\frac{1}4$

So $\rm\: d = 5a\!-\!1 = \frac{1}4,\: $ $\rm\:c = 3d = \frac{3}4,\:$ $\rm\: b = -3a\!-\!d = -\frac{3}4-\frac{1}4 = -1$

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