What is the value of $c$ in $x^{2} + 8 x + c$ that would make it a perfect square trinomial?

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Solution

Find the value of $c$ .
Recall the identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ . Here $a^{2} + 2 a b + b^{2}$ represents the perfect square of $(a + b)$ .
Compare the perfect square trinomial $a^{2} + 2 a b + b^{2}$ with the given trinomial $x^{2} + 8 x + c$ . Observe that $a = x$ , $2 a b = 8 x$ , and $b^{2} = c$ .
Substitute $a = x$ in the equation $2 a b = 8 x$ and then solve for $b$ :
$\begin{array}{rcl} 2 a b & = & 8 x \\ \Rightarrow & 2 x b = 8 x \\ \Rightarrow & b = \frac{8 x}{2 x} \\ \Rightarrow & b = 4 \end{array}$
Substitute, $b = 4$ in the equation $b^{2} = c$ and then evaluate $c$ :
$\begin{array}{rcl} b^{2} & = & c \\ \Rightarrow & 4^{2} = c \\ \Rightarrow & 16 = c \end{array}$
Therefore, the given trinomial $x^{2} + 8 x + c$ is a perfect square for $c = 16$ .

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