For what real values of \(c\) is \(4x^2 + 14x + c\)the square of a binomial?
I'd do it like this
\(4x^2 + 14x + c\\ =4(x^2+\frac{14}{4}x+\frac{c}{4})\\ =4(x^2+\frac{7}{2}x+\frac{c}{4})\\\)
now if this is a perfect square then:
\(\frac{c}{4}=[\frac{7}{2}\color{red}\div2\color{black}]\color{red}^2\\ \frac{c}{4}=[\frac{7}{4}]\color{red}^2\\ \frac{c}{4}=\frac{49}{16}\\ c=\frac{49}{4}=12\frac{1}{4}\)
LaTex:
\frac{c}{4}=[\frac{7}{2}\color{red}\div2\color{black}]\color{red}^2\\
\frac{c}{4}=[\frac{7}{4}]\color{red}^2\\
\frac{c}{4}=\frac{49}{16}\\
c=\frac{49}{4}=12\frac{1}{4}