A technical documentation specialist is writing an instructional guide for algebraic modeling in industrial design. In the section on 'completing the square' for an expression in the form $x^2 + bx$, the specialist must outline the standard rule for finding the constant that creates a perfect square trinomial. What are the two specific mathematical steps that must be performed on the linear coefficient ($b$) to determine this constant?

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**Completing the square** is an algebraic technique for determining the constant that must be added to a two-term expression like $$x^2 + 6x$$ to transform it into a perfect square trinomial — one that can be factored as the square of a binomial. The technique is motivated by the need to apply the Square Root Property, which requires the variable side of an equation to be a perfect square. When the expression is not already a perfect square, completing the square provides a way to make it one.

The method works by reversing the Binomial Squares Pattern. Since $$(a + b)^2 = a^2 + 2ab + b^2$$, the middle term $$2ab$$ is always twice the product of the binomial's two terms. Consider the expression $$x^2 + 6x$$. The first term $$x^2$$ identifies $$a = x$$. The linear term $$6x$$ corresponds to $$2ab$$ in the pattern, so $$2 \cdot x \cdot b = 6x$$, which means $$b = \frac{1}{2} \cdot 6 = 3$$. To find the missing constant:

1. Take **half** the coefficient of the linear term: $$\frac{1}{2} \cdot 6 = 3$$. This gives the value of $$b$$.
2. **Square** that result to obtain $$b^2$$: $$3^2 = 9$$.

Adding $$9$$ to the original expression produces a perfect square trinomial: $$x^2 + 6x + 9 = (x + 3)^2$$. The sign between the two original terms determines which form of the Binomial Squares Pattern applies — a positive linear term uses $$(a + b)^2$$, while a negative linear term uses $$(a - b)^2$$.

Completing the Square

To complete the square of an expression of the form $$x^2 + bx$$, follow a three-step procedure:

**Step 1.** **Identify $$b$$**, the coefficient of the linear term $$x$$.

**Step 2.** **Compute $$\left(\frac{1}{2}b\right)^2$$** — take half of $$b$$ and square the result. This value is the constant needed to complete the square.

**Step 3.** **Add $$\left(\frac{1}{2}b\right)^2$$ to $$x^2 + bx$$.** The resulting expression $$x^2 + bx + \left(\frac{1}{2}b\right)^2$$ is a perfect square trinomial that factors as $$\left(x + \frac{1}{2}b\right)^2$$.

When $$b$$ is negative, the procedure works the same way — halving a negative number produces a negative result, but squaring it always yields a positive constant. The factored binomial square then uses subtraction rather than addition.

How to Complete a Square

Complete the square for the expression $$x^2 + 14x$$ to form a perfect square trinomial, then express the result as a binomial square.

**Step 1 — Identify $$b$$.** The coefficient of $$x$$ is $$14$$, so $$b = 14$$.

**Step 2 — Find $$\left(\frac{1}{2}b\right)^2$$.** Compute half of $$14$$: $$\frac{1}{2} \cdot 14 = 7$$. Square the result: $$7^2 = 49$$.

**Step 3 — Add $$49$$ to the expression.**

$$x^2 + 14x + 49$$

**Rewrite as a binomial square.** The completed trinomial factors as:

$$x^2 + 14x + 49 = (x + 7)^2$$

This example applies the completing-the-square procedure when the linear coefficient is a positive even integer. Taking half of $$14$$ yields $$7$$, and squaring $$7$$ produces the constant $$49$$ that transforms the two-term expression into the perfect square trinomial $$(x + 7)^2$$.

Completing the Square for $$x^2 + 14x$$

Complete the square for the expression $$m^2 - 26m$$ to form a perfect square trinomial, then express the result as a binomial square.

**Step 1 — Identify $$b$$.** The coefficient of $$m$$ is $$-26$$, so $$b = -26$$.

**Step 2 — Find $$\left(\frac{1}{2}b\right)^2$$.** Compute half of $$-26$$: $$\frac{1}{2} \cdot (-26) = -13$$. Square the result: $$(-13)^2 = 169$$.

**Step 3 — Add $$169$$ to the expression.**

$$m^2 - 26m + 169$$

**Rewrite as a binomial square.** Because the linear term is negative, the factored form uses subtraction:

$$m^2 - 26m + 169 = (m - 13)^2$$

This example demonstrates how the completing-the-square procedure handles a negative linear coefficient. Halving $$-26$$ gives $$-13$$, and squaring a negative number still produces a positive constant ($$169$$). The negative sign in the original expression determines that the binomial square uses the subtraction form of the Binomial Squares Pattern: $$(m - 13)^2$$.

Completing the Square for $$m^2 - 26m$$

Complete the square for the expression $$u^2 - 9u$$ to form a perfect square trinomial, then express the result as a binomial square.

**Step 1 — Identify $$b$$.** The coefficient of $$u$$ is $$-9$$, so $$b = -9$$.

**Step 2 — Find $$\left(\frac{1}{2}b\right)^2$$.** Compute half of $$-9$$: $$\frac{1}{2} \cdot (-9) = -\frac{9}{2}$$. Square the result: $$\left(-\frac{9}{2}\right)^2 = \frac{81}{4}$$.

**Step 3 — Add $$\frac{81}{4}$$ to the expression.**

$$u^2 - 9u + \frac{81}{4}$$

**Rewrite as a binomial square.** Because the linear term is negative, the factored form uses subtraction:

$$u^2 - 9u + \frac{81}{4} = \left(u - \frac{9}{2}\right)^2$$

This example demonstrates what happens when the linear coefficient is an **odd integer**. Unlike even coefficients (such as $$14$$ or $$-26$$), taking half of an odd number produces a fraction — here, $$\frac{1}{2} \cdot (-9) = -\frac{9}{2}$$. Squaring that fraction gives another fraction: $$\left(-\frac{9}{2}\right)^2 = \frac{81}{4}$$. As a result, both the constant added to complete the square and the number inside the binomial square are fractions rather than integers. The procedure itself is identical to the integer-coefficient cases — halving a negative odd number still yields a negative result, and squaring it still produces a positive constant.

Completing the Square for $$u^2 - 9u$$

Complete the square for the expression $$p^2 + \frac{1}{2}p$$ to form a perfect square trinomial, then express the result as a binomial square.

**Step 1 — Identify $$b$$.** The coefficient of $$p$$ is $$\frac{1}{2}$$, so $$b = \frac{1}{2}$$.

**Step 2 — Find $$\left(\frac{1}{2}b\right)^2$$.** Compute half of $$\frac{1}{2}$$: $$\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$. Square the result: $$\left(\frac{1}{4}\right)^2 = \frac{1}{16}$$.

**Step 3 — Add $$\frac{1}{16}$$ to the expression.**

$$p^2 + \frac{1}{2}p + \frac{1}{16}$$

**Rewrite as a binomial square.** Because the linear term is positive, the factored form uses addition:

$$p^2 + \frac{1}{2}p + \frac{1}{16} = \left(p + \frac{1}{4}\right)^2$$

This example demonstrates the completing-the-square procedure when the linear coefficient is itself a **fraction** rather than an integer. Taking half of $$\frac{1}{2}$$ gives $$\frac{1}{4}$$, and squaring $$\frac{1}{4}$$ gives $$\frac{1}{16}$$. Each step involves multiplying fractions — $$\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$ and $$\frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}$$ — but the overall procedure remains the same three steps used for integer coefficients. The result is a perfect square trinomial whose constant term is a small fraction.

Completing the Square for $$p^2 + \frac{1}{2}p$$

To solve a quadratic equation of the form $$x^2 + bx + c = 0$$ by completing the square, follow a six-step procedure:

**Step 1.** **Isolate the variable terms on one side and the constant terms on the other.** Move the constant $$c$$ to the right side of the equation so the left side contains only $$x^2 + bx$$.

**Step 2.** **Find $$\left(\frac{1}{2} \cdot b\right)^2$$, the number needed to complete the square, and add it to both sides of the equation.** Take half the coefficient of the linear term, square it, and add that value to both sides. This preserves the equality — whatever is added to one side must also be added to the other.

**Step 3.** **Factor the perfect square trinomial as a binomial square.** The left side is now a perfect square trinomial that factors into the square of a binomial.

**Step 4.** **Use the Square Root Property.** Apply $$x + \frac{b}{2} = \pm\sqrt{k}$$, where $$k$$ is the simplified value on the right side.

**Step 5.** **Simplify the radical and solve the two resulting equations.** If the radicand is a perfect square, simplify to an integer; otherwise, simplify using the Product Property of Square Roots.

**Step 6.** **Check the solutions.** Substitute each value back into the original equation to confirm both sides are equal.

This procedure combines the completing-the-square technique (Steps 1–3) with the Square Root Property (Steps 4–5) to solve quadratic equations that may not be factorable. The critical requirement in Step 2 is that the same constant must be added to **both** sides of the equation to maintain the equality.

How to Solve a Quadratic Equation by Completing the Square

Solve $$x^2 + 8x = 48$$ by applying the six-step completing-the-square procedure.

**Step 1 — Isolate the variable terms.** The variable terms $$x^2 + 8x$$ are already on the left and the constant $$48$$ is on the right, so no rearranging is needed.

**Step 2 — Find $$\left(\frac{1}{2} \cdot b\right)^2$$ and add it to both sides.** The coefficient of $$x$$ is $$8$$, so $$b = 8$$. Compute half of $$8$$: $$\frac{1}{2} \cdot 8 = 4$$. Square the result: $$4^2 = 16$$. Add $$16$$ to both sides:

$$x^2 + 8x + 16 = 48 + 16$$

**Step 3 — Factor the perfect square trinomial.** The left side factors as a binomial square:

$$(x + 4)^2 = 64$$

**Step 4 — Apply the Square Root Property:**

$$x + 4 = \pm\sqrt{64}$$

**Step 5 — Simplify the radical and solve.** Since $$64$$ is a perfect square ($$8^2 = 64$$):

$$x + 4 = \pm 8$$

Write as two equations and solve each:

$$x + 4 = 8 \implies x = 4$$

$$x + 4 = -8 \implies x = -12$$

**Step 6 — Check both solutions:**

For $$x = 4$$: $$4^2 + 8(4) = 16 + 32 = 48$$ ✓

For $$x = -12$$: $$(-12)^2 + 8(-12) = 144 - 96 = 48$$ ✓

The solutions are $$x = 4$$ and $$x = -12$$. This example demonstrates the simplest case of solving by completing the square: the variable terms are already isolated, the coefficient of $$x$$ is a positive even integer, and the resulting constant on the right side is a perfect square, so the solutions are integers.

Solving $$x^2 + 8x = 48$$ by Completing the Square

Solve $$y^2 - 6y = 16$$ by completing the square, demonstrating the procedure when the linear coefficient is negative.

**Step 1 — Isolate the variable terms.** The variable terms are already on the left side.

**Step 2 — Find $$\left(\frac{1}{2} \cdot b\right)^2$$ and add to both sides.** The coefficient of $$y$$ is $$-6$$, so $$b = -6$$. Compute: $$\left(\frac{1}{2}(-6)\right)^2 = (-3)^2 = 9$$. Add $$9$$ to both sides:

$$y^2 - 6y + 9 = 16 + 9$$

**Step 3 — Factor the perfect square trinomial.** Because the linear term is negative, the binomial square uses subtraction:

$$(y - 3)^2 = 25$$

**Step 4 — Apply the Square Root Property:**

$$y - 3 = \pm\sqrt{25}$$

**Step 5 — Simplify and solve.** Since $$25$$ is a perfect square ($$5^2 = 25$$):

$$y - 3 = \pm 5$$

Write as two equations:

$$y - 3 = 5 \implies y = 8$$

$$y - 3 = -5 \implies y = -2$$

**Step 6 — Check both solutions:**

For $$y = 8$$: $$8^2 - 6(8) = 64 - 48 = 16$$ ✓

For $$y = -2$$: $$(-2)^2 - 6(-2) = 4 + 12 = 16$$ ✓

The solutions are $$y = 8$$ and $$y = -2$$. This example shows how completing the square works when the linear coefficient is **negative**: halving $$-6$$ gives $$-3$$, and squaring $$-3$$ still yields a positive constant ($$9$$). The negative sign in the original expression determines that the factored form uses subtraction — $$(y - 3)^2$$ rather than $$(y + 3)^2$$.

Solving $$y^2 - 6y = 16$$ by Completing the Square

Solve $$x^2 + 10x + 4 = 15$$ by completing the square, demonstrating the procedure when constants appear on both sides of the equation.

**Step 1 — Isolate the variable terms.** The variable terms $$x^2$$ and $$10x$$ are already on the left, but the constant $$4$$ must be moved to the right. Subtract $$4$$ from both sides:

$$x^2 + 10x = 11$$

**Step 2 — Find $$\left(\frac{1}{2} \cdot b\right)^2$$ and add it to both sides.** The coefficient of $$x$$ is $$10$$, so $$b = 10$$. Compute: $$\left(\frac{1}{2}(10)\right)^2 = 5^2 = 25$$. Add $$25$$ to both sides:

$$x^2 + 10x + 25 = 11 + 25$$

**Step 3 — Factor the perfect square trinomial.** The left side factors as a binomial square:

$$(x + 5)^2 = 36$$

**Step 4 — Apply the Square Root Property:**

$$x + 5 = \pm\sqrt{36}$$

**Step 5 — Simplify the radical and solve.** Since $$36$$ is a perfect square ($$6^2 = 36$$):

$$x + 5 = \pm 6$$

Write as two equations and solve each:

$$x + 5 = 6 \implies x = 1$$

$$x + 5 = -6 \implies x = -11$$

**Step 6 — Check both solutions:**

For $$x = 1$$: $$1^2 + 10(1) + 4 = 1 + 10 + 4 = 15$$ ✓

For $$x = -11$$: $$(-11)^2 + 10(-11) + 4 = 121 - 110 + 4 = 15$$ ✓

The solutions are $$x = 1$$ and $$x = -11$$. Unlike earlier examples where the equation was already in the form $$x^2 + bx = c$$, this equation has a constant on the **same side** as the variable terms. Step 1 requires subtracting that constant from both sides before the completing-the-square process can begin. The trinomial $$x^2 + 10x + 25$$ matches the Binomial Squares Pattern because $$(x + 5)^2 = x^2 + 10x + 25$$.

Solving $$x^2 + 10x + 4 = 15$$ by Completing the Square

Solve $$n^2 = 3n + 11$$ by completing the square, demonstrating the procedure when variable terms must first be collected on one side and the linear coefficient is an odd number.

**Step 1 — Collect the variable terms on one side.** Subtract $$3n$$ from both sides to move all variable terms to the left:

$$n^2 - 3n = 11$$

**Step 2 — Find $$\left(\frac{1}{2} \cdot b\right)^2$$ and add it to both sides.** The coefficient of $$n$$ is $$-3$$, so $$b = -3$$. Compute: $$\left(\frac{1}{2}(-3)\right)^2 = \left(-\frac{3}{2}\right)^2 = \frac{9}{4}$$. Add $$\frac{9}{4}$$ to both sides:

$$n^2 - 3n + \frac{9}{4} = 11 + \frac{9}{4}$$

**Step 3 — Factor the perfect square trinomial.** The left side factors as a binomial square using the subtraction form:

$$\left(n - \frac{3}{2}\right)^2 = \frac{44}{4} + \frac{9}{4} = \frac{53}{4}$$

**Step 4 — Apply the Square Root Property:**

$$n - \frac{3}{2} = \pm\sqrt{\frac{53}{4}}$$

**Step 5 — Simplify the radical and solve.** Apply the Quotient Property to the radical: $$\sqrt{\frac{53}{4}} = \frac{\sqrt{53}}{\sqrt{4}} = \frac{\sqrt{53}}{2}$$. Since $$53$$ is prime, $$\sqrt{53}$$ cannot be simplified further. Add $$\frac{3}{2}$$ to both sides:

$$n = \frac{3}{2} \pm \frac{\sqrt{53}}{2}$$

Write as two solutions:

$$n = \frac{3}{2} + \frac{\sqrt{53}}{2} \quad \text{or} \quad n = \frac{3}{2} - \frac{\sqrt{53}}{2}$$

The solutions are $$n = \frac{3 + \sqrt{53}}{2}$$ and $$n = \frac{3 - \sqrt{53}}{2}$$. This example combines two complications: first, the variable terms appear on **both sides** of the original equation, so they must be collected to one side before the procedure can begin. Second, the linear coefficient $$-3$$ is odd, so halving it produces the fraction $$-\frac{3}{2}$$, and the resulting constant $$\frac{9}{4}$$ introduces fractions throughout the remaining steps. The radicand $$53$$ is not a perfect square, so the final solutions remain in radical form.

Solving $$n^2 = 3n + 11$$ by Completing the Square

Solve $$(x - 3)(x + 5) = 9$$ by completing the square, demonstrating the procedure when the equation begins as a product of two distinct binomials equal to a nonzero constant.

**Multiply the binomials on the left.** Apply the FOIL method to expand $$(x - 3)(x + 5)$$: First $$x \cdot x = x^2$$, Outer $$x \cdot 5 = 5x$$, Inner $$(-3) \cdot x = -3x$$, Last $$(-3) \cdot 5 = -15$$. Combine like terms:

$$x^2 + 2x - 15 = 9$$

**Step 1 — Isolate the variable terms.** Add $$15$$ to both sides to move the constant away from the variable terms:

$$x^2 + 2x = 24$$

**Step 2 — Find $$\left(\frac{1}{2} \cdot b\right)^2$$ and add it to both sides.** The coefficient of $$x$$ is $$2$$, so $$b = 2$$. Compute: $$\left(\frac{1}{2}(2)\right)^2 = 1^2 = 1$$. Add $$1$$ to both sides:

$$x^2 + 2x + 1 = 24 + 1$$

**Step 3 — Factor the perfect square trinomial.** The left side factors as a binomial square:

$$(x + 1)^2 = 25$$

**Step 4 — Apply the Square Root Property:**

$$x + 1 = \pm\sqrt{25}$$

**Step 5 — Simplify and solve.** Since $$25$$ is a perfect square ($$5^2 = 25$$):

$$x + 1 = \pm 5$$

Write as two equations and solve each:

$$x + 1 = 5 \implies x = 4$$

$$x + 1 = -5 \implies x = -6$$

The solutions are $$x = 4$$ and $$x = -6$$. This example introduces an additional preliminary step not present in earlier completing-the-square problems: the equation is given as a **product of two different binomials** equal to a constant, so the binomials must first be multiplied out using FOIL before the standard six-step procedure can begin. The expanded form $$x^2 + 2x - 15 = 9$$ then has constants on both sides, requiring the constant $$-15$$ to be moved to the right before completing the square.

Solving $$(x - 3)(x + 5) = 9$$ by Completing the Square

Solve $$3x^2 - 12x - 15 = 0$$ by completing the square, demonstrating the procedure when the leading coefficient is not $$1$$ but can be factored out as a GCF.

**Preliminary step — Make the leading coefficient $$1$$.** The coefficient of $$x^2$$ is $$3$$, which divides evenly into all three terms. Factor out $$3$$:

$$3(x^2 - 4x - 5) = 0$$

Divide both sides by $$3$$:

$$x^2 - 4x - 5 = 0$$

**Step 1 — Isolate the variable terms.** Add $$5$$ to both sides to move the constant to the right:

$$x^2 - 4x = 5$$

**Step 2 — Find $$\left(\frac{1}{2} \cdot b\right)^2$$ and add it to both sides.** The coefficient of $$x$$ is $$-4$$, so $$b = -4$$. Compute: $$\left(\frac{1}{2}(-4)\right)^2 = (-2)^2 = 4$$. Add $$4$$ to both sides:

$$x^2 - 4x + 4 = 5 + 4$$

**Step 3 — Factor the perfect square trinomial.** The left side factors as a binomial square using the subtraction form:

$$(x - 2)^2 = 9$$

**Step 4 — Apply the Square Root Property:**

$$x - 2 = \pm\sqrt{9}$$

**Step 5 — Simplify and solve.** Since $$9$$ is a perfect square ($$3^2 = 9$$):

$$x - 2 = \pm 3$$

Write as two equations and solve each:

$$x - 2 = 3 \implies x = 5$$

$$x - 2 = -3 \implies x = -1$$

**Step 6 — Check both solutions:**

For $$x = 5$$: $$3(5)^2 - 12(5) - 15 = 75 - 60 - 15 = 0$$ ✓

For $$x = -1$$: $$3(-1)^2 - 12(-1) - 15 = 3 + 12 - 15 = 0$$ ✓

The solutions are $$x = 5$$ and $$x = -1$$. This example introduces a preliminary step not present in earlier completing-the-square problems: the leading coefficient $$3$$ must be removed before the standard procedure can begin. Because $$3$$ divides evenly into all three coefficients ($$3$$, $$-12$$, and $$-15$$), factoring it out and dividing both sides by $$3$$ reduces the equation to the familiar form $$x^2 + bx + c = 0$$ with a leading coefficient of $$1$$.

Solving $$3x^2 - 12x - 15 = 0$$ by Completing the Square

A landscape architect is designing a square garden plot. To complete the square for the expression x^2 + 10x, arrange the following steps in the correct order.

A floor installer is calculating the area for a square room. To transform the expression x^2 + 14x into a perfect square trinomial, what constant value must be added?

A logistics coordinator is modeling the expansion of a square shipping zone using the expression x^2 + 20x. To complete the square and determine the necessary constant to add to the expression, the coordinator must take half of the linear coefficient (20) and then ____ that value.

An inventory specialist is designing a series of square storage bins for a new warehouse layout. To calculate the necessary dimensions using the 'completing the square' method, match each algebraic layout expression with the specific constant value that must be added to transform it into a perfect square trinomial.

Standard Procedure for Completing the Square

True or False: A facilities manager calculating the dimensions for a square office expansion using the expression $x^2 + 16x$ determines that the constant value to be added to 'complete the square' is found by first squaring the linear coefficient (16) and then taking half of that result.

Based on the provided audit scenario, identify and state the two-step mathematical rule that the technician must recall to correctly determine the completion constant for the design expression.

Structural Integrity Audit: Square Completion Protocol

You are training a new quality control technician on algebraic optimization techniques used in facility layout planning. Write a brief explanation describing the standard two-step procedure used to 'complete the square' for a quadratic expression like $x^2 + bx$. Your response should identify exactly what operations must be performed on the linear coefficient to determine the specific constant that transforms the expression into a perfect square trinomial.

Documentation of the 'Completing the Square' Algorithm

A technical documentation specialist is updating a manual for an engineering firm. In a section regarding quadratic modeling, they describe adding a constant to an expression of the form $x^2 + bx$ so that it can be factored as the square of a binomial. What is the specific algebraic name for the resulting three-term expression?

A technical illustrator is creating a procedural flowchart for the 'completing the square' method using the expression $x^2 - 12x$. When documenting the step where a constant is added to create a perfect square trinomial, which of the following fundamental rules must the illustrator recall?

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