Completing the square is a method to solve quadratic equations by transforming one side into a perfect square trinomial. This technique is fundamental and is even used to derive the quadratic formula.
Understanding the Goal: The aim is to convert a quadratic equation from the standard `ax² + bx + c = 0` form to `(x + a)² = constant` or `(x - a)² = constant`. This allows you to easily solve for `x` by taking the square root of both sides. Step 1: Isolate the x² and x terms: Move the constant term to the right side of the equation. Step 2: Ensure the coefficient of x² is 1: If the `x²` term has a coefficient other than 1, divide the entire equation by that coefficient. Step 3: Complete the Square: Step 4: Rewrite and Solve:
<example> Recall that `(x+a)² = x² + 2ax + a²`. We want to make our equation look like the right side of this identity on one side. </example>
<example> For `x² + 16x - 57 = 0`, add 57 to both sides to get `x² + 16x = 57`. </example>
<common-mistake> A common mistake is to forget this step when the `x²` coefficient is not 1. For `6x² - 7x - 3 = 0`, first move the constant: `6x² - 7x = 3`. Then, divide by 6: `x² - (7/6)x = 1/2`. </common-mistake>
Take the coefficient of the `x` term.
Divide it by 2.
Square the result.
Add this squared value to both sides of the equation. This addition "completes" the perfect square trinomial on the left side.
<tip> If you have `x² + 2ax`, you need to add `a²` to complete the square. The `x` term's coefficient is `2a`, so `a` is half of that coefficient. Squaring `a` gives you the term to add.
</tip>
<example> In `x² + 16x = 57`, the coefficient of `x` is 16. Half of 16 is 8. Squaring 8 gives 64. So, add 64 to both sides: `x² + 16x + 64 = 57 + 64`. </example>
Rewrite the left side as a squared expression, e.g., `(x + a)²` or `(x - a)²`.
Simplify the right side.
Take the square root of both sides, remembering to include both the positive and negative roots (`±`).
Solve for `x`.
<example> From `(x + 8)² = 121`, take the square root of both sides: `x + 8 = ±√121` which simplifies to `x + 8 = ±11`. Finally, solve for `x`: `x = -8 ± 11`. This gives two solutions: `x = 3` and `x = -19`. </example>
Completing the square is a powerful method for solving quadratic equations, especially when factoring isn't straightforward or when the solutions are not whole numbers. It also forms the basis for deriving the quadratic formula.
The core idea behind completing the square is to manipulate a standard quadratic equation ($ax^2 + bx + c = 0$) into the form $(x+a)^2 = \text{number}$. Once in this form, you can easily solve for $x$ by taking the square root of both sides.
What is a perfect square trinomial?
A perfect square trinomial is a trinomial that results from squaring a binomial.
<example>
$(x+a)^2 = x^2 + 2ax + a^2$
$(x-a)^2 = x^2 - 2ax + a^2$
</example>
Notice the pattern: the constant term ($a^2$) is the square of half the coefficient of the $x$ term ($2a$).
Let's break down the process with examples.
Consider the equation: $x^2 + 16x - 57 = 0$
1. Isolate the $x^2$ and $x$ terms: Move the constant term to the right side of the equation.
Add 57 to both sides:
$x^2 + 16x = 57$
2. Find the term to "complete the square":
Focus on the coefficient of the $x$ term. In this case, it's 16.
Divide this coefficient by 2: $16 / 2 = 8$.
Square the result: $8^2 = 64$. This is the number you need to add to both sides of the equation to make the left side a perfect square trinomial.
<common-mistake>
A common mistake is forgetting to add the term to both sides of the equation. This violates the fundamental rule of algebra for maintaining equality.
Incorrect: $x^2 + 16x + 64 = 57$
Correct: $x^2 + 16x + 64 = 57 + 64$
</common-mistake>
3. Add the calculated term to both sides:
$x^2 + 16x + 64 = 57 + 64$
$x^2 + 16x + 64 = 121$
4. Rewrite the left side as a squared binomial:
The left side is now a perfect square trinomial. It can be written as $(x + \text{half of the x-coefficient})^2$.
$(x + 8)^2 = 121$
5. Take the square root of both sides: Remember to include both the positive and negative square roots on the right side.
<tip>
Always remember the "plus or minus" $(\pm)$ when taking the square root of both sides of an equation! Forgetting this will result in losing one of the solutions.
</tip>
$\sqrt{(x+8)^2} = \pm\sqrt{121}$
$x + 8 = \pm 11$
6. Solve for $x$: This will give you two possible solutions.
Case 1: $x + 8 = 11$
$x = 11 - 8$
$x = 3$
Case 2: $x + 8 = -11$
$x = -11 - 8$
$x = -19$
So the solutions are $x = 3$ and $x = -19$.
Consider the equation: $6x^2 - 7x - 3 = 0$
1. Isolate the $x^2$ and $x$ terms: Move the constant term to the right side.
$6x^2 - 7x = 3$
2. Make the coefficient of $x^2$ equal to 1: Divide every term in the equation by the coefficient of $x^2$.
<common-mistake>
Forgetting to divide all terms (including the constant on the right side) by the coefficient of $x^2$ is a common error.
Incorrect: $x^2 - 7x/6 = 3$ (If the right side was not divided)
Correct: $\frac{6x^2}{6} - \frac{7x}{6} = \frac{3}{6}$
</common-mistake>
$x^2 - \frac{7}{6}x = \frac{1}{2}$
3. Find the term to "complete the square":
Take the coefficient of the $x$ term: $-\frac{7}{6}$.
Divide it by 2 (or multiply by $\frac{1}{2}$): $-\frac{7}{6} \times \frac{1}{2} = -\frac{7}{12}$.
Square the result: $\left(-\frac{7}{12}\right)^2 = \frac{49}{144}$.
So, $\frac{49}{144}$ is the number to add to both sides.
4. Add the calculated term to both sides:
$x^2 - \frac{7}{6}x + \frac{49}{144} = \frac{1}{2} + \frac{49}{144}$
5. Rewrite the left side as a squared binomial:
$(x - \frac{7}{12})^2 = \frac{1}{2} + \frac{49}{144}$
<tip>
To add fractions on the right side, find a common denominator. The common denominator for 2 and 144 is 144.
$\frac{1}{2} = \frac{1 \times 72}{2 \times 72} = \frac{72}{144}$
</tip>
$(x - \frac{7}{12})^2 = \frac{72}{144} + \frac{49}{144}$
$(x - \frac{7}{12})^2 = \frac{121}{144}$
6. Take the square root of both sides: Remember the $\pm$.
$\sqrt{\left(x - \frac{7}{12}\right)^2} = \pm\sqrt{\frac{121}{144}}$
$x - \frac{7}{12} = \pm \frac{11}{12}$
7. Solve for $x$:
Case 1: $x - \frac{7}{12} = \frac{11}{12}$
$x = \frac{11}{12} + \frac{7}{12}$
$x = \frac{18}{12} = \frac{3}{2}$
Case 2: $x - \frac{7}{12} = -\frac{11}{12}$
$x = -\frac{11}{12} + \frac{7}{12}$
$x = -\frac{4}{12} = -\frac{1}{3}$
The solutions are $x = \frac{3}{2}$ and $x = -\frac{1}{3}$.
The term "completing the square" comes from the action of adding a specific constant term to a binomial expression ($x^2 + bx$) to turn it into a perfect square trinomial ($x^2 + bx + (b/2)^2$), which can then be factored into $(x + b/2)^2$. You are literally "completing" the algebraic "square" form on one side of the equation.
The quadratic formula ($x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$) is actually derived by applying the completing the square method to the general quadratic equation $ax^2 + bx + c = 0$. This highlights the fundamental importance of completing the square in algebra.