Example 2
Completion requirements
| Example 2 |
Find the roots of \(x^2 + 10x + 4 = 0\) by completing the square.
Step 1: Start by converting the equation to vertex form by completing the square.
\(\begin{align}
x^2 + 10x + 4 &= 0 \\
\left( {x^2 + 10x} \right) + 4 &= 0 \\
\left( {x^2 + 10x + 5^2 - 5^2 } \right) + 4 &= 0 \\
\left( {x^2 + 10x + 5^2 } \right) + 4 - 25 &= 0 \\
\left( {x + 5} \right)^2 - 21 &= 0 \end{align}\)
Step 2: Isolate the squared binomial, then take the square root of both sides.
\(\begin{align}
\left( {x + 5} \right)^2 - 21 &= 0 \\
\left( {x + 5} \right)^2 &= 21 \\
\sqrt {\left( {x + 5} \right)^2} &= \pm \sqrt {21} \\
x + 5 &= \pm \sqrt {21} \end{align}\)
Step 3: Solve for \(x\).
The question does not ask for rounded solutions, so leave them as exact values.
\(\begin{align}
x^2 + 10x + 4 &= 0 \\
\left( {x^2 + 10x} \right) + 4 &= 0 \\
\left( {x^2 + 10x + 5^2 - 5^2 } \right) + 4 &= 0 \\
\left( {x^2 + 10x + 5^2 } \right) + 4 - 25 &= 0 \\
\left( {x + 5} \right)^2 - 21 &= 0 \end{align}\)
Step 2: Isolate the squared binomial, then take the square root of both sides.
\(\begin{align}
\left( {x + 5} \right)^2 - 21 &= 0 \\
\left( {x + 5} \right)^2 &= 21 \\
\sqrt {\left( {x + 5} \right)^2} &= \pm \sqrt {21} \\
x + 5 &= \pm \sqrt {21} \end{align}\)
Step 3: Solve for \(x\).
\(\begin{align}
x + 5 &= \pm \sqrt {21} \\
x &= -5 \pm \sqrt {21} \\
x &= -5 + \sqrt {21} \thinspace {\rm{ and }} \thinspace x = -5 - \sqrt {21} \end{align}\)
x + 5 &= \pm \sqrt {21} \\
x &= -5 \pm \sqrt {21} \\
x &= -5 + \sqrt {21} \thinspace {\rm{ and }} \thinspace x = -5 - \sqrt {21} \end{align}\)
The question does not ask for rounded solutions, so leave them as exact values.