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Concept Version 10
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Completing the Square

Completing the square is a method for solving quadratic equations, and involves putting the quadratic in the form $0=a(x-h)^2 + k$.

Learning Objective

  • Solve for the zeros of a quadratic function by completing the square


Key Points

    • Quadratics can be put in the form $a(x-h)^2 + k$, where $h$ and $k$ are constants specific to a given quadratic.
    • Once a quadratic polynomial is in the form: $0=a(x-h)^2+k$ , one can solve for two values of $x$ (using the positive and negative square roots).

Term

  • quadratic equation

    A polynomial equation of the second degree.


Full Text

Along with factoring and using the quadratic formula, completing the square is a common method for solving quadratic equations.  It is often implemented when factoring is not an option, such as when the quadratic is a not already a perfect square.

Consider the formula for a generic quadratic equation:

$0=ax^2+bx+c$

The method of completing the square allows for the conversion to the form:

$0=a(x-h)^2+k$

where $h$ and $k$ are constants with a specific value. 

The value of $k$ is meant to adjust the function to compensate for the difference between the expanded form of $a(x-h)^2$ and the general quadratic function $ax^2+bx+c$.  This adjustment is what mathematically allows for the two forms to be equal.

Once completing the square has been performed, the quadratic is easy to solve; because there is only one place where the variable $x$ is squared, the $(x-h)^2$ term can be isolated on one side of the equation, and then the square root of both sides can be taken.

Example

As an example, consider the following quadratic polynomial:

${ x }^{ 2 }+10x+22$

This quadratic is not a perfect square.  The closest perfect square is the square of $5$, which was determined by dividing the $b$ term (in this case $10$) by two and producing the square of the result.  

${ { (x+5) }^{ 2 }=x }^{ 2 }+10x+25$

However, it is possible to write the original quadratic as the sum of this square and a constant:

$\begin{aligned} { x }^{ 2 }+10x+22 &= x^2 + 10x + 25 - 3 \\ &= (x+5)^2 - 3 \end{aligned}$

Thus, the constant $h$ takes the value $-5$ and the constant $k$ takes the value $-3$.

Knowing this, we can now solve for x:

$\begin{aligned} (x+5)^2-3& =0 \\ (x+5)^2 &=3 \\ x+5&=\pm \sqrt 3\\ x&= \pm \sqrt 3 -5\end{aligned}$

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