Binomial

What number must be placed in the box?

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\(9x^2 + 6x + c=0\)

a          b        c

\(x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\ The\ radicant\ must\ be\ Zero\ for\ x\ to\ have\ only\ one\ value.\\ b^2-4ac=0\\ c=\dfrac{b^2}{4a}=\dfrac{6^2}{4\cdot 9}\\ \color{blue}c=1\)

\(x=-\dfrac{b}{2a}=-\dfrac{6}{2\cdot 9}\\ \color{blue} x=-\dfrac{1}{3}\)

\((x+\frac{1}{3})^2=x^2+\frac{2}{3}x+\frac{1}{9}=(3x+1)^2=9x^2+6x+1=0\)

1 must be placed in the box.

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