banner
home icon
 
Related Lessons: Solving Quadratic Equations by Quadratic Formula Solving Quadratic Equations by Completing-the-Square Method Deriving the Quadratic Formula

 

Completing the Square | Step by Step

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
y=x^2-4x+3 See Solution
y=2x^2+6x-1 See Solution
y=-3x^2+3x+2 You Try
y=5x^2+15x-5 You Try

 

Type 2: Solve the quadratic equation using Completing the Square
x^2-9x+14=0 See Solution
x^2+8x+2=22 You Try
6x^2+69x-36=0 See Solution
-3x^2-2x+5=-3 You Try

 


Example 1: Find the vertex form of the quadratic function y=x^2-4x+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.

showing the coefficient of x-term in y=x^2-4x+3

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

(-4/2)^2=(-2)^2=4

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

boy thinking 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.

y=(x^2-4x+4)+3-4

 

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, , then the parabola opens in upward direction.

 

shows the graph of quadratic function with vertex at the point (2,-1). This is a minimum value.

 

 


Example 2: Find the vertex form of the quadratic function y=2x^2+6x-1.

The approach to this problem is slightly different because the value of "a" doesn't equal 1, "a" does not equal to 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+3x)-1

 

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

shows the coefficient of the x-term in y=2(x^2+3x)-1

 

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

(3/2)^2 = 9/4

 

STEP 4: Now, I will take the output 9/4 and add it inside the parenthesis.

By adding 9/4 inside the parenthesis, I am actually adding 2(9/4)=9/2 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.

y=2[x^2+3x+(9/4)]-1
STEP 5: Since I added 9/2 to the equation, then I should subtract the entire equation by 9/2 also to compensate for it.

 

y=2[x^2+3x+(9/4)]-1-(9/2)

 

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)62+k where the vertex (h,k) is the vertex at (-3/2,-11/2).

y=2[x+(3/2)]^2-(11/2)

Example 3: Find the vertex form of the quadratic function y=-3x^2+3x+2.

Try this problem yourself on paper. Click here to show the answer in a pop-up window.

 

 


Example 4: Find the vertex form of the quadratic function y=5x^2+15x-5.

Try this problem yourself on paper. Click here to show the answer in pop-up window.

 

 

 


Example 5: Solve the quadratic equation x^2-9x+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.

x^2-9x+14-14=0-14, x^2-9x=-14

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

shows the coefficient of the linear term: x^2-9x=-14

Take that number, divide by 2 and square it.

(-9/2)^2=81/4

 

Add 81/4 to both sides of the equation, and then simplify.

x^2-9x+81/4 = 25/4

 

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

[x-(9/2)]^2=25/4

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).

 

x-(9/2)=± (5/2)

Solve for "x" by adding both sides by 9/2.

x=±(5/2) + (9/2)

Find the two values of "x" by considering the two cases: positive and negative.

shows the steps to solve for the two values of x: x1 = 7 and x2=2

 

Therefore, the final answers are x1= 7 and x2= 2. You may back-substitute these two values of x from the original equation to check.


Example 6: Solve the quadratic equation x^2+8x+2=22 using Completing the Square method.

Try this out on paper and click here to compare your solution. The answer will open in a pop-up window.

 


Example 7: Solve the quadratic equation 6x^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.

6x^2+69x-36+36=0+36, 6x^2+69x=36

Divide the entire equation by the coefficient of the x2 term which is 6. Reduce the fraction to its lowest term.

x^2+(23/2)x=6

Identify the coefficient of the linear term.

shows the coefficient of the x-term of x^2+(23/2)=6

Divide this coefficient by 2 and square it.

[(23/2)/2]^2 = (23/4)^2 = 529/16

 

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

 

x^2+(23/2)x+(529/16)=625/16

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.

[x+(23/4)]^2=625/16, x+(23/4)=± 25/4

 

Finish this off by subtracting both sides by 23/4. You should obtain two values of "x" because of the "plus or minus".

shows the two values of x: x1 = 1/2, and x2 = -12

The final answers are x1= 1/2 and x2= −12.

 


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

Try this first on paper. Whenever you're ready to compare your solution, please click here. The solution will open in a pop-up window.

 


 

Take a Quiz!
About This Topic: Completing the Square

 

Practice Problems with Answers
Worksheet 1 Worksheet 2

 

Top of Page
Custom Search
JUMP TO:
Algebra Lessons
Algebra Lessons (A-Z)
Algebra Worksheets
Free Math Solver
Find Local Math Tutors
Math Games
 

 

 

HOME   |   PRIVACY POLICY   |   CONTACT US   |   DONATE

© Copyright 2011-2014   ChiliMath   All Rights Reserved