Completing the Square: Videos & Practice Problems

Completing the square transforms any quadratic polynomial into a perfect square trinomial, enabling solutions via the square root property. By representing $x^2$ as a square and splitting the linear term's coefficient, students learn to add or subtract constants to balance equations. This method applies to polynomials of degree two, involving terms with exponents and coefficients, and is essential for solving quadratics not in standard form. Understanding this process deepens comprehension of binomials, monomials, and the structure of polynomials, enhancing problem-solving skills in algebra.

Complete the Square

Completing the square is a versatile method for solving any quadratic equation by transforming it into a perfect square trinomial. This technique involves rewriting a quadratic expression in the form ax² + bx + c into a squared binomial plus or minus a constant, allowing the use of the square root property to find solutions.

To understand completing the square visually, consider the quadratic expression x² + 6x + 12. The term x² can be represented as the area of a square with side length x. The linear term 6x is split into two equal parts, 3x and 3x, each represented as rectangles adjoining the square. By arranging these shapes, a larger square with side length x + 3 is nearly formed, except for a small missing square in the corner with area 3 × 3 = 9. This means x² + 6x + 9 can be expressed as (x + 3)². Since the original quadratic has a constant term of 12, which is 3 more than 9, we add 3 to balance the equation, resulting in (x + 3)² + 3. This process of creating a perfect square trinomial is the essence of completing the square.

Applying this method to solve equations, take the quadratic x² + 2x - 8 = 0. First, represent x² as a square and split the middle term 2x into two parts of x each. These correspond to rectangles with side lengths x and 1, forming a nearly complete square with side length x + 1. The missing corner square has an area of 1, so x² + 2x + 1 equals (x + 1)². Since the original constant is -8, which is 9 less than 1, subtract 9 from both sides to maintain equality: (x + 1)² - 9 = 0. Adding 9 to both sides gives (x + 1)² = 9. Taking the square root of both sides yields x + 1 = ±\sqrt{9}, simplifying to x + 1 = ±3. Solving for x gives two solutions: x = 2 and x = -4.

The key formula in completing the square involves adding and subtracting the square of half the coefficient of x. For a quadratic in the form x² + bx + c, the perfect square trinomial is (x + \frac{b}{2})² = x² + bx + \left(\frac{b}{2}\right)². Adjusting the equation by adding or subtracting the difference between c and \left(\frac{b}{2}\right)² allows the quadratic to be rewritten as a perfect square plus a constant.

Completing the square not only provides a reliable method for solving any quadratic equation but also deepens understanding of the structure of quadratics. By transforming the equation into a perfect square form, it becomes straightforward to apply the square root property and find exact solutions. This method is especially useful when factoring is difficult or impossible, and it lays the foundation for deriving the quadratic formula and analyzing the properties of parabolas.

Determine the value we need to add to the equation to make it a perfect square trinomial, then factor it.
$x^2+8x+$__

• A. $16;\left(x+4\right)^2$
• B. $8;\left(x+4\right)^2$
• C. $16;\left(x+2\right)^2$
• D. $8;\left(x+2\right)^2$

Answer: (A)

Determine the value we need to add to the equation to make it a perfect square trinomial, then factor it.
$y^2-10y+$ __

• A. $5;\left(y-5\right)^2$
• B. $25;\left(y+5\right)^2$
• C. $25;\left(y-5\right)^2$
• D. $5;\left(y+5\right)^2$

Answer: (C)

Complete the Square Example 1

When solving quadratic equations by completing the square, special attention is needed when the leading coefficient a is not one or when the linear coefficient b is an odd number. The method can still be applied effectively by first manipulating the equation into a more manageable form.

For a quadratic equation such as \$2x^2 - 4x - 1 = 0\(, where the leading coefficient \)a = 2\(, the first step is to divide the entire equation by 2 to normalize the coefficient of \)x^2\( to 1. This yields:

\[x^2 - 2x - \frac{1}{2} = 0\]

Next, to complete the square, focus on the \)x^2 - 2x\( portion. The coefficient of \)x\( is \)-2\(, so half of this value is \)-1\(. Squaring \)-1\( gives \)1\(, which is the number needed to complete the square. Adding and subtracting 1 inside the equation allows rewriting the quadratic as a perfect square trinomial:

\[x^2 - 2x + 1 - 1 - \frac{1}{2} = 0 \implies (x - 1)^2 - \frac{3}{2} = 0\]

Isolating the perfect square term and solving for \)x\( involves adding \)\frac{3}{2}\( to both sides and then applying the square root property:

\[(x - 1)^2 = \frac{3}{2}\]\[x - 1 = \pm \sqrt{\frac{3}{2}}\]\[x = 1 \pm \sqrt{\frac{3}{2}}\]

This results in two solutions for \)x\( expressed in simplest radical form.

In cases where the linear coefficient \)b\( is odd, such as in the equation \)z^2 - 3z - 4 = 0\(, the process begins by ensuring the equation is in standard form. Then, half of \)b\( is taken, which may be fractional. Here, half of \)-3\( is \)-\frac{3}{2}\(. Squaring this value gives \)\left(-\frac{3}{2}\right)^2 = \frac{9}{4}\(. Adding and subtracting \)\frac{9}{4}\( completes the square:

\[z^2 - 3z + \frac{9}{4} - \frac{9}{4} - 4 = 0 \implies \left(z - \frac{3}{2}\right)^2 - \frac{25}{4} = 0\]

Isolating the perfect square term and solving for \)z\( involves adding \)\frac{25}{4}$ to both sides and applying the square root property:

\[\left(z - \frac{3}{2}\right)^2 = \frac{25}{4}\]\[z - \frac{3}{2} = \pm \frac{5}{2}\]\[z = \frac{3}{2} \pm \frac{5}{2}\]

Evaluating these expressions yields the two solutions:

\[z = 4 \quad \text{or} \quad z = -1\]

Completing the square is a versatile method that transforms quadratic equations into perfect square trinomials, enabling straightforward application of the square root property. This approach is especially useful when factoring is difficult or when the quadratic formula is not preferred. Key steps include normalizing the leading coefficient to one, halving the linear coefficient, squaring it to find the constant term needed to complete the square, and carefully balancing the equation by adding or subtracting the same value on both sides. This method reinforces understanding of quadratic structure and provides exact solutions in radical or simplified fractional form.

Solve for $x$
$8x^2-20x+12=4$

• A. $x=3,x=-\frac12$
• B. $x=3,x=\frac12$
• C. $x=2,x=-\frac12$
• D. $x=2,x=\frac12$

Answer: (D)

Here’s what students ask on this topic:

What is the method of completing the square in solving quadratic equations?

Completing the square is a method used to solve any quadratic equation by transforming it into a perfect square trinomial. This involves rewriting the quadratic expression in the form $x^2+\; bx\; +\; c$ as ${x+d}^2+\; e$, where $d$ and $e$ are constants. The process starts by representing $x^2$ as a square and splitting the middle term $bx$ into two equal parts. Then, you add a constant to complete the square, balancing the equation by adding or subtracting the same value on both sides. This allows you to apply the square root property to solve for $x$. Completing the square works for all quadratic equations and helps deepen understanding of polynomial structure and algebraic manipulation.

How do you complete the square for the quadratic expression $x^2\; +\; 6x\; +\; 12$?

To complete the square for $x^2\; +\; 6x\; +\; 12$, first focus on the $x^2\; +\; 6x$ part. Split the middle term coefficient 6 in half to get 3. Then, square 3 to get 9. Add and subtract 9 inside the expression to keep it balanced: $x^2\; +\; 6x\; +\; 9\; -\; 9\; +\; 12$. Now, $x^2\; +\; 6x\; +\; 9$ is a perfect square trinomial and can be written as $(x\; +\; 3)^2$. So, the expression becomes $(x\; +\; 3)^2\; +\; 3$. This form makes it easier to solve equations or analyze the quadratic by using the square root property.

Why is completing the square considered a universal method for solving quadratic equations?

Completing the square is considered universal because it can be applied to any quadratic equation, regardless of its coefficients. Unlike factoring, which only works when the quadratic factors nicely, completing the square always transforms the quadratic into a perfect square trinomial. This allows the use of the square root property to find solutions. The method involves algebraic manipulation that balances the equation by adding or subtracting constants, ensuring the equation remains equivalent. This universality makes it a powerful tool for solving quadratics, especially when other methods fail or are complicated.

How do you solve the quadratic equation $x^2\; +\; 2x\; -\; 8\; =\; 0$ by completing the square?

To solve $x^2\; +\; 2x\; -\; 8\; =\; 0$ by completing the square, first move the constant to the other side: $x^2\; +\; 2x\; =\; 8$. Next, take half of the coefficient of $x$, which is 2, so half is 1. Square it to get 1. Add 1 to both sides: $x^2\; +\; 2x\; +\; 1\; =\; 8\; +\; 1$, simplifying to $(x\; +\; 1)^2\; =\; 9$. Now apply the square root property: $x\; +\; 1\; =\; B1\; 3$. Finally, solve for $x$: $x\; =\; -1\; B1\; 3$, giving two solutions: $x\; =\; 2$ and $x\; =\; -4$.

What is the significance of the constant term when completing the square?

The constant term plays a crucial role in completing the square because it determines how the quadratic expression is balanced. When you create a perfect square trinomial by adding a specific constant (the square of half the middle term's coefficient), you must adjust the original constant term to keep the equation equivalent. This often involves adding or subtracting a value on both sides of the equation. Properly managing the constant term ensures the quadratic can be rewritten as a binomial squared plus or minus a constant, which is essential for applying the square root property and solving the equation accurately.