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 x² + bx + c as a squared binomial plus or minus a constant, allowing the use of the square root property to find solutions.

To understand completing the square, visualize x² as the area of a square with side length x. The linear term, such as 6x, can be split into two equal parts (3x + 3x) and represented as two rectangles adjoining the square. By adding a smaller square in the corner with side length equal to half the coefficient of x (in this case, 3), you complete a larger square with side length x + 3. The area of this larger square is expressed algebraically as \((x + 3)^2 = x^2 + 6x + 9\).

When the original quadratic has a constant term different from this perfect square (for example, 12 instead of 9), adjust the equation by adding or subtracting the difference on both sides to maintain balance. This process converts the quadratic into the form \((x + p)^2 = q\), where p and q are constants.

For example, to solve \(x^2 + 2x - 8 = 0\) by completing the square, first rewrite the quadratic as \((x + 1)^2 - 9 = 0\) because \((x + 1)^2 = x^2 + 2x + 1\). Then, add 9 to both sides to isolate the square: \((x + 1)^2 = 9\). Applying the square root property gives \(x + 1 = \pm \sqrt{9}\), simplifying to \(x + 1 = \pm 3\). Solving for x yields two solutions: \(x = -1 + 3 = 2\) and \(x = -1 - 3 = -4\).

This method not only provides a systematic way to solve any quadratic equation but also deepens understanding of the structure of quadratics by linking algebraic expressions to geometric representations. The key formula used in completing the square is:

\[x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2\]

where \(\frac{b}{2}\) is half the coefficient of the linear term. By adding and subtracting \(\left(\frac{b}{2}\right)^2\) appropriately, the quadratic is rewritten as a perfect square trinomial plus or minus a constant, enabling straightforward solutions through square roots.

Completing the square is essential for solving quadratics that are not easily factorable and serves as a foundation for deriving the quadratic formula. It also enhances problem-solving skills by encouraging manipulation of equations into more manageable forms and verifying solutions by substitution back into the original equation.

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

• A. $16;{(x+4)}^2$
• B. $8;{(x+4)}^2$
• 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 where the coefficient of x² is not one, such as \$2x^2 - 4x - 1 = 0\(, the first step is to divide the entire equation by the leading coefficient to normalize the quadratic term. Dividing by 2 yields:

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

Next, to complete the square, focus on the coefficient of the linear term, which is now -2. Half of this value is -1, and squaring it gives 1. This means the expression can be rewritten as a perfect square trinomial:

\[x^2 - 2x + 1 = (x - 1)^2\]

However, since the original equation has \)-\frac{1}{2}\( instead of \)+1\(, adjust by subtracting the difference \)\left(1 - \frac{1}{2} = \frac{3}{2}\right)\( from both sides:

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

Solving for \)x\( involves isolating the squared term and 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 yields two solutions for \)x\(.

When the linear coefficient \)b\( is an odd number, such as in the equation \)z^2 - 3z - 4 = 0\(, the process is similar but requires careful handling of fractions. First, ensure the equation is in standard form. Then, take half of the linear coefficient \)-3\(, which is \)-\frac{3}{2}\(, and square it:

\[\left(-\frac{3}{2}\right)^2 = \frac{9}{4}\]

This allows rewriting the quadratic as a perfect square trinomial:

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

Since the original constant term is \)-4\(, adjust by subtracting the difference between \)-4\( and \)\frac{9}{4}\(. Converting \)-4\( to quarters gives \)-\frac{16}{4}$, so the difference is:

\[-\frac{16}{4} - \frac{9}{4} = -\frac{25}{4}\]

Rewrite the equation as:

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

Isolate the perfect square and solve:

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

This results in two solutions:

\[z = \frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4\]\[z = \frac{3}{2} - \frac{5}{2} = -\frac{2}{2} = -1\]

Completing the square is a powerful technique for solving quadratic equations, especially when factoring is difficult or impossible. By normalizing the quadratic term and carefully handling fractional values, any quadratic equation can be transformed into a perfect square trinomial, making it straightforward to solve using the square root property.

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)

Complete the Square Example 2

To solve a quadratic equation by completing the square, start by visualizing the quadratic expression as a geometric square diagram. The term x² represents the area of a large square with side lengths x. The linear term, such as -2x, is split into two equal parts, each represented as rectangles with one side length x and the other side length being half the coefficient of x, in this case, -1. Completing the square involves adding the area of a smaller square formed by these half-lengths, which is calculated by squaring the half coefficient: \((-1) \times (-1) = 1\).

This process transforms the quadratic expression into a perfect square trinomial: \(x^2 - 2x + 1 = (x - 1)^2\). However, if the original quadratic has a constant term different from this added square, adjust the equation accordingly. For example, if the original quadratic is \(x^2 - 2x + 3\), rewrite it as \((x - 1)^2 + 2\) by adding 2 to both sides to maintain equality.

Setting the equation equal to zero allows solving for x. For instance, \((x - 1)^2 + 2 = 0\) leads to \((x - 1)^2 = -2\). Applying the square root property, take the square root of both sides: \(x - 1 = \pm \sqrt{-2}\). Recognizing that the square root of a negative number involves imaginary numbers, express \(\sqrt{-2}\) as \(\sqrt{2} \times \sqrt{-1} = \sqrt{2}i\), where i is the imaginary unit defined by \(i^2 = -1\).

Finally, solve for x by isolating it: \(x = 1 \pm \sqrt{2}i\). This solution demonstrates how completing the square not only simplifies quadratic expressions but also reveals complex roots when the quadratic does not intersect the x-axis, highlighting the importance of imaginary numbers in solving quadratic equations.

Here’s what students ask on this topic:

What is the method of completing the square in algebra?

Completing the square is a method used to solve quadratic equations by transforming any quadratic polynomial into a perfect square trinomial. This process involves rewriting the quadratic expression in the form $x^2+\; bx\; +\; c$ as ${x+d}^2+\; e$, where $d$ and $e$ are constants. The key step is to take half of the coefficient of the linear term ($b$), square it, and add it inside the expression to form a perfect square trinomial. This allows the quadratic to be expressed as a binomial squared, making it easier to solve using the square root property. Completing the square works for any quadratic equation and helps visualize the structure of the polynomial.

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 $x$ terms. Take half of the coefficient of $x$, which is 6, so half is 3. Then square 3 to get 9. Rewrite the expression as $x^2+\; 6x\; +\; 9\; +\; (12\; -\; 9)$. This becomes $(x\; +\; 3)^2\; +\; 3$. Here, $(x\; +\; 3)^2$ is the perfect square trinomial, and the constant 3 balances the equation. This form makes it easier to solve or analyze the quadratic by using the square root property.

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

Completing the square is useful because it transforms any quadratic equation into a form that can be solved using the square root property. Unlike factoring, which only works when the quadratic factors nicely, completing the square works for all quadratics. It also helps in understanding the vertex form of a quadratic function, which is important for graphing and analyzing the parabola's properties. By rewriting the quadratic as a perfect square trinomial plus or minus a constant, you can isolate the squared term and take square roots to find solutions efficiently.

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

Start with the equation $x^2+\; 2x\; -\; 8\; =\; 0$. Move the constant to the other side: $x^2+\; 2x\; =\; 8$. Take half of the coefficient of $x$, which is 2, so half is 1, and square it to get 1. Add 1 to both sides: $x^2+\; 2x\; +\; 1\; =\; 8\; +\; 1$. This gives $(x\; +\; 1)^2\; =\; 9$. Take the square root of both sides: $x\; +\; 1\; =\; B1\; 3$. Solve for $x$: $x\; =\; -1\; B1\; 3$, so $x\; =\; 2$ or $x\; =\; -4$. These are the solutions.

What is the visual interpretation of completing the square?

Visually, completing the square can be understood by representing $x^2$ as the area of a square with side length $x$. The linear term, such as $6x$, is split into two rectangles each with side length $x$ and the other side length half of the coefficient (e.g., 3). These rectangles are arranged around the square to form an almost complete larger square. The missing small square in the corner has an area equal to the square of half the linear coefficient (e.g., 9). Adding this small square completes the larger square, which corresponds algebraically to forming a perfect square trinomial. This geometric approach helps students understand why the method works.