**Completing the Square**

In a regular algebra class, completing the square is usually used to either find the vertex form of a quadratic function (parabola) or solve a quadratic equation.

In this lesson, I will go over a few examples that would illustrate the usefulness of completing the square in these two types of problems.

**TYPE 1: Find the Vertex Form using Completing the Square**

**Example 1**: Find the vertex form of the quadratic function **y = x**^{2}** – ****4****x + ****3**.

This quadratic equation is in the form **y = ax**^{2}* + bx + c*. However, I need to rewrite it using some algebraic steps in order to make it look like this…

*y = a(x – h)*^{2}* + k*

This is the vertex form of the quadratic function where (*h,k*) is the vertex or the “center” of the quadratic function or the parabola.

Before I start, I realize that *a*=1. Therefore, I can immediately apply the “completing the square” steps.

**STEP 1:** Identify the coefficient of the linear term of the quadratic function. That is the number attached to the x-term.

**STEP 2:** I will take that number, divide it by 2 and square it (or raise to the power 2).

**STEP 3:** The output in step #2 will be added and subtracted on the same side of the equation to keep it balanced.

Think about it…if I add 4 on the right side of the equation, then I am technically changing the original meaning of the equation. So to keep it unchanged, I must subtract the same value that I added on the same side of the equation.

**STEP 4:** Now, express the trinomial inside the parenthesis as a square of a binomial, and simplify the outside constants.

After simplifying, it is now in the vertex form *y = a(x – h)*^{2}* + k* where the vertex (*h,k*) is (2, –1).

*y = (x – *2*) ^{2} – *1

Visually, the graph of this quadratic function is a parabola with a minimum at the point (2,−1). Since the value of “*a*” is positive, *a*=1, then the parabola opens in upward direction.

**Example 2**: Find the vertex form of the quadratic function **y = ****2****x**^{2}** + ****6****x – ****1**.

The approach to this problem is slightly different because the value of “*a*” doesn’t equal 1, *a *≠ 1. The first step is to factor out the coefficient 2 between the terms with x-variables only.

**STEP 1:** Factor out 2 only to the terms with variable *x*.

*y = *2(*x ^{2} + *3

*x) –*1

**STEP 2:** Identify the coefficient of the *x*-term or linear term.

**STEP 3:** Take that number, divide it by 2, and square.

**STEP 4:** Now, I will take the output and add it inside the parenthesis.

By adding inside the parenthesis, I am actually adding to the entire equation.

Why multiply by 2 to get the “true” value added to the entire equation? Remember, I factored out 2 in the beginning. So for us to find the real value added to the entire equation, we need to multiply the number added inside the parenthesis by the number that was factored out.

**STEP 5:** Since I added to the equation, then I should subtract the entire equation by also to compensate for it.

**STEP 6:** Finally, express the trinomial inside the parenthesis as the square of binomial and then simplify the outside constants. Be careful combining the fractions.

It is now in the vertex form *y = a*(*x – h**)*^{2}* + k* where the vertex **( h,k)** is .

**Example 3**: Find the vertex form of the quadratic function *y = –*3*x*^{2}* + *3*x** + ***2**.

Try this problem yourself on paper. Then click “**Solution**” to compare your answer.

Factor out -3 among the x-terms. The coefficient of the linear term inside the parenthesis is -1. Divide it by 2 and square it. Add that value inside the parenthesis. Now, figure out how to make the original equation the same. Since we added 1/4 inside the parenthesis and we factored out -3 in the beginning, that means -3 (1/4) = -3/4 is the value that we subtracted from the entire equation. To compensate, we must add 3/4 outside the parenthesis.

Therefore, the vertex (*h,k*) is .

**Example 4**: Find the vertex form of the quadratic function *y = *5*x*^{2}* + *15*x** – ***5**.

Try this problem yourself on paper. Then click “**Solution**” to compare your answer.

Factor out 5 among the x-terms. Identify the coefficient of the linear term inside the parenthesis which is 3. Divide it by 2 and square to get 9/4.

Add 9/4 inside the parenthesis. Since we factored out 5 in the first step, that means 5 (9/4) = 45/4 is the number that we need to subtract to keep the equation unchanged.

Express the trinomial as a square of binomial, and combine the constants to get the final answer.

Therefore, the vertex (*h,k*) is .

**TYPE 2: Solve the Quadratic Equation using Completing the Square**

**Example 5**: Solve the quadratic equation **x**^{2}** – ****9****x + ****14**** = ****0** using Completing the Square method.

Move the constant to the right side of the equation, while keeping the *x*-terms on the left. I can do that by subtracting both sides by 14.

Next, identify the coefficient of the linear term (just the *x*-term) which is

Take that number, divide by 2 and square it.

Add to both sides of the equation, and then simplify.

Express the trinomial on the left side as a square of binomial.

Take the square roots of both sides of the equation to eliminate the power of 2 of the parenthesis. Make sure that you attach the plus or minus symbol to the constant term (right side of equation).

Solve for “*x*” by adding both sides by .

Find the two values of “*x*” by considering the two cases: positive and negative.

Therefore, the final answers are ** x_{1}= 7** and

**. You may back-substitute these two values of**

*x*_{2}= 2*x*from the original equation to check.

**Example 6**: Solve the quadratic equation ** x^{2} + 8x + 2 = 22**.

Try this problem yourself on paper. Then click “**Solution**” to compare your answer.

**Example 7**: Solve the quadratic equation **6 x^{2} + 69x – 36 = 0** using Completing the Square method.

Eliminate the constant −36 on the left side by adding 36 to both sides of the quadratic equation.

Divide the entire equation by the coefficient of the *x*^{2} term which is 6. Reduce the fraction to its lowest term.

Identify the coefficient of the linear term.

Divide this coefficient by 2 and square it.

Add this output to both sides of the equation. Be careful when adding or subtracting fractions.

Express the trinomial on the left side as a perfect square binomial. Then solve the equation by first taking the square roots of both sides. Don’t forget to attach the** plus or minus symbol** to the square root of the constant term on the right side.

Finish this off by subtracting both sides by . You should obtain two values of “*x*” because of the “plus or minus”.

The final answers are ** x_{1}= 1/2** and

**.**

*x*_{2}= −12

**Example 8**: Solve the quadratic equation ** –3x^{2} – 2x + 5 = -3** using Completing the Square method .

Try this first on paper. Then click “**Solution**” to compare your answer.

**Step 1:** Eliminate the constant on the left side, and then divide the entire equation by −3.

**Step 2:** Take the coefficient of the linear term which is 2/3. Divide it by 2 and square it.

**Step 3:** Add the value found in step #2 to both sides of the equation. Then combine the fractions.

**Step 4:** Express the trinomial on the left side as square of a binomial.

**Step 5:** Take the square roots of both sides of the equation. Make sure that you attach the “plus or minus” symbol to the square root of the constant on the right side. Simplify the radical.

**Step 6:** Solve for *x* by subtracting both sides by 1/3. You should have two answers because of the “plus or minus” case.