“Completing the square” is a powerful technique in algebra that transforms a quadratic equation into a more manageable form. It might seem a bit abstract at first, but it’s a fundamental skill that unlocks deeper understanding and problem-solving abilities in mathematics.
| … start with a Quadratic Equation like this: | … and convert it into this format: |
|---|---|
| ax2 + bx + c = 0 | a(x+d)2 + e = 0 |
For those seeking a quick formula, here are the values for d and e:
d = b/(2a)
and
e = c − (b2)/(4a)
However, to truly grasp the power and application of completing the square, let’s delve into the “how” and “why” behind this method. Understanding the process will empower you to use it effectively in various mathematical contexts.
Understanding the Concept of Completing the Square
Let’s begin with a simple algebraic expression: x2 + bx. Having ‘x’ appear twice can sometimes complicate things. Our goal with completing the square is to rewrite this expression in a way that ‘x’ appears only once, simplifying it for further manipulation or problem-solving.
Visualizing this geometrically can offer a helpful insight. Imagine x2 as the area of a square and bx as the area of a rectangle. We can arrange these shapes to almost form a larger square:
As illustrated, x2 + bx is nearly a square. To “complete the square“, we need to add a small square piece, which turns out to be (b/2)2.
Algebraically, this transformation looks like this:
| x2 + bx | + (b/2)2 | = | (x+b/2)2 |
|---|---|---|---|
| “Completing the Square” |
By adding (b/2)2, we successfully “complete the square,” resulting in an expression where ‘x’ appears only once: (x+b/2)2. This form is often much easier to work with in various algebraic manipulations and equation solving.
Maintaining Equation Balance: Adding and Subtracting
Now, a crucial point to remember in mathematics: we can’t simply add a term to an expression without maintaining the overall value. To add (b/2)2 and keep the expression equivalent to the original, we must also subtract it. This ensures we are only rewriting the expression, not changing its value.
Let’s illustrate this with a numerical example to clarify the process:
| Start with: | x2 + 6x + 7 |
|---|---|
| (“b” is 6 in this case) | |
| Complete the Square: | x2 + 6x + (6/2)2 + 7 – (6/2)2 (Add and subtract the new term) |
| Simplify it and we are done. |
Simplifying the expression:
x2 + 6x + 7 = (x2 + 6x + 9) + 7 – 9 = (x+3)2 − 2
The result:
x2 + 6x + 7 = (x+3)2 − 2
As you can see, we successfully completed the square, and now ‘x’ appears only once in the form (x+3)2 − 2.
Shortcut Method for Completing the Square
For those who prefer a more direct approach, here’s a shortcut method that becomes quite efficient with practice. We aim to transform our expression into the form: (x+d)2 + e.
Expanding (x+d)2 gives us: x2 + 2dx + d2 + e
Now, let’s take our previous example and fit it into this expanded form to determine the values of ‘d’ and ‘e’.
Example: Fitting x2 + 6x + 7 into x2 + 2dx + d2 + e
By comparing the coefficients, we can deduce:
- The term 6x must correspond to 2dx. Therefore, 2d = 6, which means d = 3.
- The constant term 7 must correspond to d2 + e. Since d = 3, d2 = 9. So, 7 = 9 + e, which means e = -2.
Substituting d = 3 and e = -2 into (x+d)2 + e, we get (x+3)2 − 2, which is the same result we obtained earlier. This shortcut method offers a quicker way to complete the square once you become comfortable with it.
Solving Quadratic Equations by Completing the Square
One of the most significant applications of completing the square is solving quadratic equations. Recall that a general quadratic equation is in the form ax2 + bx + c = 0. Completing the square provides a systematic method to find the solutions (roots) of such equations.
However, if there’s a coefficient ‘a’ in front of x2 (where a ≠ 1), we first need to divide the entire equation by ‘a’ to simplify the process:
x2 + (b/a)x + c/a = 0
Step-by-Step Guide to Solving Quadratic Equations by Completing the Square
Here are the five key steps to solve a quadratic equation using the completing the square method:
- Step 1: Divide by ‘a’ (if necessary). If the coefficient of x2 (which is ‘a’) is not 1, divide every term in the equation by ‘a’.
- Step 2: Isolate the x2 and x terms. Move the constant term (c/a) to the right side of the equation.
- Step 3: Complete the square. Take half of the coefficient of the x term (which is b/a, so half is b/2a), square it ((b/2a)2), and add this value to both sides of the equation. This step completes the square on the left side.
- Step 4: Take the square root. The left side of the equation is now a perfect square trinomial, which can be written in the form (x + p)2. Take the square root of both sides of the equation. Remember to consider both positive and negative square roots.
- Step 5: Solve for x. Isolate ‘x’ by subtracting the constant term from both sides of the equation.
Examples of Solving Quadratic Equations
Let’s work through a couple of examples to solidify your understanding.
Example 1: Solve x2 + 4x + 1 = 0
Step 1: The coefficient of x2 is already 1, so we can skip this step.
Step 2: Move the constant term to the right side:
x2 + 4x = -1
Step 3: Complete the square. Half of the coefficient of x is (4/2) = 2, and squaring it gives 22 = 4. Add 4 to both sides:
x2 + 4x + 4 = -1 + 4
(x + 2)2 = 3
Step 4: Take the square root of both sides:
x + 2 = ±√3 ≈ ±1.73 (rounded to two decimal places)
Step 5: Solve for x by subtracting 2 from both sides:
x = ±1.73 – 2
This gives us two solutions:
x ≈ -1.73 – 2 = -3.73
x ≈ 1.73 – 2 = -0.27
| Interestingly, at the end of Step 3, we had the equation: (x + 2)2 = 3. This form directly reveals the vertex (turning point) of the parabola represented by x2 + 4x + 1, which is (-2, -3). | |
Example 2: Solve 5x2 – 4x – 2 = 0
Step 1: Divide all terms by 5:
x2 – 0.8x – 0.4 = 0
Step 2: Move the constant term to the right side:
x2 – 0.8x = 0.4
Step 3: Complete the square. Half of the coefficient of x is (-0.8/2) = -0.4, and squaring it gives (-0.4)2 = 0.16. Add 0.16 to both sides:
x2 – 0.8x + 0.16 = 0.4 + 0.16
(x – 0.4)2 = 0.56
Step 4: Take the square root of both sides:
x – 0.4 = ±√0.56 ≈ ±0.748 (rounded to three decimal places)
Step 5: Solve for x by adding 0.4 to both sides:
x = ±0.748 + 0.4
This yields two solutions:
x ≈ -0.748 + 0.4 = -0.348
x ≈ 0.748 + 0.4 = 1.148
Why is Completing the Square Important?
You might wonder, “Why bother with completing the square when we have the Quadratic Formula to solve quadratic equations?”. While the quadratic formula is certainly a handy tool, completing the square offers several advantages:
- Reveals the Vertex: As seen in Example 1, the completed square form directly shows the vertex of the parabola, which is crucial in graphing quadratic functions and understanding their properties.
- Foundation for the Quadratic Formula: Completing the square is actually the method used to derive the quadratic formula itself. Understanding completing the square gives you a deeper appreciation of where the quadratic formula comes from.
- Simplifies Complex Problems: In more advanced mathematical contexts, expressions in the form ax2 + bx + c might be part of larger, more complex problems. Rewriting them in the completed square form, a(x+d)2 + e, can significantly simplify these problems because ‘x’ appears only once, making further manipulations easier. For instance, if ‘x’ is itself a function like cos(z), completing the square can pave the way for a more elegant solution.
- Enhances Algebraic Understanding: Mastering completing the square strengthens your algebraic manipulation skills and provides a more profound understanding of quadratic expressions and equations.
Think of completing the square as another essential tool in your mathematical toolbox. It provides not just a method for solving equations, but also a deeper insight into the structure and properties of quadratic expressions.
Footnote: Deriving the Formulas for “d” and “e”
Let’s revisit how we obtained the formulas for ‘d’ and ‘e’ at the beginning of this guide:
| Start with the general form | **** |
|---|---|
| Divide the equation by a | |
| Isolate the x2 and x terms | **** |
| Add (b/2a)2 to both sides to complete the square | |
| “Complete the Square” form | **** |
| Rearrange to bring everything back to the left side… | |
| … and multiply back by the original factor a | **** |
By comparing this final form, a(x+b/2a)2 + (c – b2/4a) = 0, with our target form a(x+d)2 + e = 0, we can clearly see that:
d = b/(2a)
and
e = c − (b2)/(4a)
These are the formulas presented at the start, derived directly from the process of completing the square.
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