How to determine the value of k of the trinomial that is a perfect square? 9x^2 + kx + 49

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How to determine the value of k of the trinomial that is a perfect square? 9x^2 + kx + 49

Algebra
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Recall that \((3x + 7)^2 = 9x^2 + \cdots+49\).
If \(ax^2+bx+c\) is a perfect square then \(b=2*\sqrt a*\sqrt c\) here a=9 and c=49 Does that help @peekaboopork
(a + b)^2 = a^2 + 2ab + b^2 9x^2 = (3x)^2 49 = 7^2 (3x + 7)^2 = 9x^2 +42x + 49 (3x - 7)^2 = 9x^2 - 42x + 49 k = 42 or k = -42

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Other answers:

A good way to find the middle term is to realize that in a perfect square, \(2ab\) is the middle term where \((a +b)^2 = a^2 + 2ab + b^2\). Here, \(b^2= 49 \iff b = 7\) and \(a^2 = 9x^2 \iff a = 3x \).
So what would \(2\cdot 7\cdot 3x\) be?
Actually I should have put \(\pm\), but it still would have the same result!

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