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What is the value of c such that: x2 + 14x + c, is a perfect-square trinomial? A. 7 B. 98 C. 196 D. 49

Algebra
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D
\[x^2+14x+c=0\]
Observe that a perfect square is in the form \((a + b)^2 = a^2 + 2ab + b^2\). The first term is \(x^2 \) so we get an idea that the perfect square is in the form \((x + b)^2\). Now the middle term is \(2ab\) and here it is \(14x\). We know that \(a = x\) and so \(2x b = 14x \Rightarrow b = 7\). And so we have \((x + 7)^2 = x^2 + 14x + 49\). There's another technique which you can use formally, which is called "Completing the Square". It's just a rip-off of what I did above.

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That technique says that you get the half of the coefficient of \(x\), and square it to get the last term. So you get the half of \(14\) which is \(7\). The square of that is \(49\).
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ok it is easier to see it like that
I have another question if you don't mind
Graph the set of points. Which model is most appropriate for the set? ( 1, 20), (0, 10), (1, 5), (2, 2.5)

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