Completing the Square: Videos & Practice Problems

Completing the square transforms any quadratic polynomial into a perfect square trinomial, enabling the use of the square root property to solve equations. By representing $x^2$ as a square and splitting the linear term's coefficient, students learn to form a binomial squared expression. Adjusting constants balances the equation, revealing solutions through square roots. This method applies to polynomials of degree two, emphasizing understanding of terms, coefficients, and exponents. Mastery of completing the square enhances problem-solving skills for quadratic equations in standard form, reinforcing concepts like monomials, trinomials, and polynomial structure.

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 taking the coefficient of the linear term, dividing it by 2, and squaring it to find the number to add and subtract inside the equation. This creates a perfect square trinomial, which can be factored into a binomial squared. Then, by isolating the squared term and applying the square root property, you can solve for $x$. This method 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 $6x$ term. Divide 6 by 2 to get 3, then square it to get 9. Rewrite the expression as $x^2+\; 6x\; +\; 9\; +\; (12\; -\; 9)$, which is $(x\; +\; 3)^2\; +\; 3$. Here, $(x\; +\; 3)^2$ is the perfect square trinomial, and the +3 adjusts the constant term to keep the expression equivalent. 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, or the quadratic formula, which is a direct formula, completing the square transforms the quadratic into a perfect square trinomial. This allows the use of the square root property to solve for $x$. It also provides insight into the structure of quadratics and is foundational for understanding more advanced topics like conic sections and quadratic functions' vertex form.

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, divide by 2 to get 1, and square it to get 1. Add 1 to both sides: $x^2+\; 2x\; +\; 1\; =\; 8\; +\; 1$. This gives $(x\; +\; 1)^2\; =\; 9$. Taking the square root of both sides yields $x\; +\; 1\; =\; B1\; 3$. Finally, solve for $x$: $x\; =\; -1\; B1\; 3$, so $x\; =\; 2$ or $x\; =\; -4$.

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

The constant term added when completing the square is crucial because it creates a perfect square trinomial from the original quadratic expression. This term is found by taking half of the coefficient of the linear term and squaring it. Adding this constant inside the expression allows the quadratic to be factored into a binomial squared, which simplifies solving the equation using the square root property. To maintain equality, the same constant must be subtracted or added outside the perfect square, ensuring the equation remains balanced. This step is essential for transforming the quadratic into a solvable form.