Skip to content# Completing the Square: Comprehensive Guide With Worksheet

## What is Completing the Square used for?

## Turning Point from Completing the Square

### Practice Questions

**Solution:**^{2 }− 6x + 9) - (9 - 9)

^{2 }+ 4x + 4) - (4 - 4)## Exploring Completing the Square with 'a' Not Equal to 1

### Practice Questions

**Solution:**^{2 }− 4x + 4 - 4) + 10

^{2 }− 4x + 4 - 4) - 6## Finding Maxima and Minima from Completing the Square

### Practice Questions

**Solution:****2. Express in vertex form using Completing the Square:****3. Rearrange the equation to isolate the perfect square:****4. Simplify the grouped terms inside the parentheses:****5. Now, the equation is in vertex form, a(x − h)**^{2} + k:^{2 }+ 6x + 9 is -9.

**2. Express in vertex form using Completing the Square:****3. Rearrange the equation to isolate the perfect square:****4. Simplify the grouped terms inside the parentheses:****5. Now, the equation is in vertex form, a(x − h)**^{2} + k:^{2 }- 12x - 15 is -51.

These are the maximum and minimum values for the respective quadratic equations.## Solving Quadratic Equations

### Practice Questions

**Solution:****2.** Express it in standard form by rearranging the terms:**3.** Divide the entire equation by 2 to simplify:**4.** To complete the square, add and subtract (b/2)^{2} inside the parentheses:**5.** Rearrange the equation by grouping the terms:**6.** Simplify the grouped terms inside the parentheses:**7.** Add 4 to both sides to isolate the perfect square:**8.** Take the square root of both sides:**9.** Solve for 'x' by subtracting 1 from both sides:

**2.** Express it in standard form by rearranging the terms:**3. **To complete the square, add and subtract (b/2)^{2} inside the parentheses:**4.** Rearrange the equation by grouping the terms:**6.** Simplify the grouped terms inside the parentheses:**7.** Add 4 to both sides to isolate the perfect square:**8.** Take the square root of both sides:**9.** Solve for 'x' by subtracting 4 from both sides:## Completing the Square GCSE questions

## Worksheet on Completing the Squares

**Completing the Square**

**What is Completing the Squares?**

- Completing the squares is a powerful technique used in algebra
**to solve quadratic equations.**

In this article, we will discuss:

**What is Completing the Squares?****What are the different applications of Completing the Squares?****GCSE Past Paper Questions**

Here is one more link to practice a few extra questions: Maths Genie Completing the Squares Questions

- Completing the Square involves converting any quadratic equation of the type:

**ax ^{2} + bx + c —-> a(x-h)^{2} + k**

It can be used to:

**Find the Turning Point of a Quadratic Equation****Find the Maximum or Minimum value of a Quadratic Equation****Solve a quadratic equation and find its roots**

**What is Turning Point of a Quadratic Equation?**

- The graph of quadratic equation represents a parabola, which can look something like this: curving either upward or downward

Our goal is to locate the **turning point of this parabola.**

*To illustrate the process, let’s work through a straightforward example:*

** **

** **

**Solved Example**

Question: Find the Turning Point of the Quadratic Equation x^{2} + 3x – 10

Solution:

**Step #1:**Begin with a quadratic equation in standard form, ax^{2}+ bx + c.

For our example, **a = 1, b = 3, and c = -10.**

**Step #2:**Apply Completing the Square to transform it into vertex form.

1. First, factor out the leading coefficient ‘a’ from the first two terms:

**x ^{2} + 3x – 10 = 1(x^{2} + 3x) – 10**

2. To complete the square, we need to add and subtract (b/2)^{2} inside the

parentheses. In this case,

**(3/2) ^{2} = 9/4. x^{2} + 3x + 9/4 – 10 – 9/4**

3. Rearrange the equation by grouping the terms:

**(x ^{2} + 3x + 9/4) – (40/4 + 9/4)**

4. Simplify both sides:

**(x + 3/2) ^{2} – 49/4**

**Step #3:**Rewrite it in vertex form, a(x-h)^{2}+ k.

In our example, a = 1, h = -3/2 (opposite sign), and k = 49/4.

So, the turning point of the quadratic equation x^{2} + 3x – 10 is (-3/2, 49/4).

Question 1. Find the turning point of the quadratic equation: **x ^{2}− 6x + 9**

To find the turning point, we'll complete the square:

**1.** Factor out the leading coefficient 'a' from the first two terms:

x^{2 }− 6x + 9 = 1(x^{2 }− 6x) + 9

**2.** Add and subtract (b/2)^{2 } inside the parentheses. In this case, (6/2)^{2 }= 9:

x^{2 }− 6x + 9 - 9 + 9

**3.** Rearrange the equation by grouping the terms:

**4.** Simplify both sides:

(x - 3)^{2} + 0

Now, we have it in vertex form, (x - h)^{2} + k, where h = 3 and k = 0.

**So, the turning point of the quadratic equation x ^{2 }− 6x + 9 is (3, 0).**

Question 2: Determine the turning point of the quadratic equation: **x ^{2 }+ 4x − 4**

**Solution:**

To find the turning point, we'll complete the square:

**1.** Factor out the leading coefficient 'a' from the first two terms:

x^{2 }+ 4x - 4 = 1(x^{2 }+ 4x) - 4

**2.** Add and subtract (b/2)^{2 } inside the parentheses. In this case, (4/2)^{2 }= 4:

x^{2 }+ 4x + 4 - 4 - 4

**3.** Rearrange the equation by grouping the terms:

**4.** Simplify both sides:

(x + 2)^{2} + 0

Now, we have it in vertex form, (x - h)^{2} + k, where h = -2 and k = 0.

So, the turning point of the quadratic equation x^{2 }+ 4x - 4 is (-2, 0).

**What happens when the coefficient of x ^{2} is not 1?**

- Up until now, we’ve worked with quadratic equations where ‘a’ (the coefficient of x
^{2}) was conveniently set to 1.

This simplified the Completing the Square process.

However, quadratic equations in real-world scenarios often have ‘a’ values other than 1.

**Let’s tackle a more complex example:**

** **

** **

**Challenging Example**

Question: Find the Turning Point of 2x^{2} – 8x – 5

Solution:

**Step #1:**Begin with the given quadratic equation:

**2x ^{2} – 8x – 5**

**Step #2:**To complete the square, factor out the leading coefficient ‘a’ (which is 2 in this case) from the first two terms:

**2(x ^{2} – 4x) – 5**

**Step #3:**Now, we need to add and subtract (b/2)^{2}inside the parentheses. Here, (4/2)^{2}=4, so we add and subtract 4:

**2(x ^{2} – 4x + 4 – 4) – 5**

**Step #4:**Rearrange the equation by grouping the terms:

**2((x ^{2} – 4x + 4) – 4) – 5**

**Step #5:**Simplify the grouped terms inside the parentheses:

**2((x – 2) ^{2} – 4) – 5**

**Step #6:**Distribute the 2 back into the expression:

**2(x – 2) ^{2} – 8 – 5**

**Step #7:**Simplify further:**2(x – 2)**^{2}– 13

Now, we have successfully completed the square, and the equation is in vertex form, a(x – h)^{2} + k, where a = 2, h = 2, and k = -13

So, the turning point of the quadratic equation 2x^{2} – 8x – 5 is (2, -13).

Question 1. Find the turning point of the quadratic equation: **3x ^{2} − 12x + 10**

To find the turning point, we'll complete the square:

**1. **Begin with the given quadratic equation:

3x^{2 }− 12x + 10

**2.** Factor out the leading coefficient 'a' (which is 3 in this case) from the first two terms:

3(x^{2 }− 4x) + 10

**3.** Add and subtract (b/2)^{2 } inside the parentheses. Here, (4/2)^{2} = 4, so we add and

subtract 4:

3(x**4.** Rearrange the equation by grouping the terms:

3((x^{2 }− 4x + 4) -4) + 10

**5.** Simplify the grouped terms inside the parentheses:

3(x - 2)^{2 }− 12 + 10

**6. **Distribute the 3 back into the expression:

3(x - 2)^{2 }− 2

Now, we have it in vertex form, a(x - h)^{2} + k, where a = 3, h = 2, and k = -2.

**So, the turning point of the quadratic equation 3x ^{2} - 12x + 10 is (2, -2).**

Question 2: Determine the turning point of the quadratic equation: **-2x ^{2} + 8x - 6**

**Solution:**

To find the turning point, we'll complete the square:

**1. **Begin with the given quadratic equation:

-2x^{2 }+ 8x - 6

**2.** Factor out the leading coefficient 'a' (which is -2 in this case) from the first two terms:

-2(x^{2 }− 4x) - 6

**3.** Add and subtract (b/2)^{2 } inside the parentheses. Here, (4/2)^{2} = 4, so we add and

subtract 4:

-2(x**4.** Rearrange the equation by grouping the terms:

-2((x^{2 }− 4x + 4) -4) - 6

**5.** Simplify the grouped terms inside the parentheses:

-2(x - 2)^{2 }+ 8 - 6

**6. **Distribute the -2 back into the expression:

-2(x - 2)^{2 }+ 2

Now, we have it in vertex form, a(x - h)^{2} + k, where a = -2, h = 2, and k = 2.

**So, the turning point of the quadratic equation -2x ^{2} + 8x - 6 is (2, 2).**

**How to find the Maximina and Minima of a Quadratic Equation?**

- The concept of
in the quadratic function represents the**maximum and minimum values**formed by the quadratic equation.**highest and lowest points on the parabolic graph**

**Maximum**values occur when the parabola opens downward**(a < 0)**,

**Minimum**values occur when it opens upward**(a > 0).**

** **

** **

**Solved Example**

Question: Find the Maximum or Minimum Value of the Quadratic Equation 2x^{2} – 8x + 6

Solution:

**Step #1:**Begin with the given quadratic equation:

**2x ^{2} – 8x + 6**

**Step #2:**Factor out the leading coefficient ‘a’ (which is 2 in this case) from the first two terms:

**2(x ^{2} – 4x) + 6**

**Step #3:**Add and subtract (b/2)^{2}inside the parentheses. Here, (4/2)^{2 }= 4, so we add and subtract 4:

**2(x ^{2} – 4x + 4 – 4) + 6**

**Step #4:**Rearrange the equation by grouping the terms:

**2((x ^{2} – 4x + 4) – 4) + 6**

**Step #5:**Simplify the grouped terms inside the parentheses:

**2(x – 2) ^{2} – 8 + 6**

**Step #6:**Distribute the 2 back into the expression:

**2(x – 2) ^{2} – 2**

Now, we have it in vertex form, a(x – h)^{2} + k, where a = 2, h = 2, and k = -2.

In this case, ‘a’ is positive (a > 0), indicating that the parabola opens upward. Therefore, the vertex represents the minimum value.

So, the minimum value of the quadratic equation 2x^{2} – 8x + 6 is -2, and it occurs at the vertex, which is (2, -2).

Question 1. Find the maximum or minimum value of the quadratic equation: **x ^{2} + 6x + 9**

**1. Identify 'a' and 'b':**

- 'a' (coefficient of x
^{2}) = 1 - 'b' (coefficient of 'x') = 6

- Add and subtract (b/2)
^{2}:

x^{2 }+ 6x + (6/2)^{2 }- (6/2)^{2} + 9

x^{2 }+ 6x + 9 - 9 + 9

(x^{2 }+ 6x + 9) - 9 = 9 - 9

(x + 3)^{2} − 9 = 0.

- 'a' is positive (a > 0), so it's a minimum.
- 'k' is the value -9.

Question 2: Determine the maximum or minimum value of the quadratic equation:** -3x ^{2} - 12x - 15**

**Solution:**

**1. Identify 'a' and 'b':**

- 'a' (coefficient of x
^{2}) = -3 - 'b' (coefficient of 'x') = -12

- Add and subtract (b/2)
^{2}:

-3x^{2 }- 12x + (-12/2)^{2 }- (-12/2)^{2} - 15

-3x^{2 }- 12x + 36 - 36 - 15

(-3x^{2 }- 12x + 36) - 36 = -15 - 36

(-3(x^{2} + 4x + 12)) − 36 = -51

- 'a' is negative (a < 0), so it's a minimum.
- 'k' is the value -51.

These are the maximum and minimum values for the respective quadratic equations.

- Completing the squares can also be used to find the roots of a quadratic equation.
- Other techniques that can be used is factorisation and quadratic formula.
- To illustrate how Completing the Square can be used for solving quadratic equations, let’s work through an example:

** **

** **

**Solved Example**

Question: Solve the Quadratic Equation x^{2} – 6x + 9 = 0

Solution:

**Step #1:**Begin with the given quadratic equation:

**x ^{2} – 6x + 9 = 0**

**Step #2:**Express it in standard form, with one side equal to zero:

**x ^{2} – 6x + 9 – 9 = 0 – 9**

**Step #3:**Factor out the leading coefficient ‘a’ (which is 1 in this case) from the first two terms:

**(x ^{2} – 6x) + 9 – 9 = -9**

**Step #4:**Add and subtract (b/2)^{2}inside the parentheses. Here, (6/2)^{2 }= 9, so we add and subtract 9:

**(x ^{2} – 6x + 9 – 9) = -9**

**Step #5:**Rearrange the equation by grouping the terms:

**(x ^{2} – 6x + 9) – 9 = -9**

**Step #6:**Simplify the grouped terms inside the parentheses:

**(x – 3) ^{2} – 9 = -9**

**Step #7:**Add 9 to both sides to isolate the perfect square:

**(x – 3) ^{2} = 0**

**Step #8:**Take the square root of both sides:

**x – 3 = ±√0**

**Step #9:**Simplify the square root of 0 to 0:

**x – 3 = ±0**

**Step #10:**Solve for ‘x’ by adding 3 to both sides:

**x = 3**

So, the solution to the quadratic equation x^{2} – 6x + 9 = 0 is x = 3.

Question 1. Solve the quadratic equation: **2x ^{2} + 4x - 6 = 0**

**1.** Begin with the given quadratic equation:

**2x ^{2 }+ 4x - 6 = 0**

**2x ^{2 }+ 4x - 6 = 0**

**x ^{2 }+ 2x - 3 = 0**

**x ^{2 }+ 2x + (2/2)^{2} - (2/2)^{2} - 3 = 0**

**(x ^{2 }+ 2x + 1) - 1 - 3 = 0**

**(x + 1) ^{2 }- 4 = 0**

**(x + 1) ^{2} = 4**

** x + 1 = ±2**

**x = −1 ± 2**

So, the solutions to the quadratic equation 2x^{2 }+ 4x - 6 = 0 are:

**x _{1} = -1 + 2 = 1**

**x _{2} = -1 – 2 = -3**

Question 2: Find the solutions to the quadratic equation: **x ^{2} + 8x + 16 = 0**

**Solution:**

**1.** Begin with the given quadratic equation:

**x ^{2 }+ 8x + 16 = 0**

**x ^{2 }+ 8x + 16 = 0**

**x ^{2 }+ 8x + (8/2)^{2} - (8/2)^{2} + 16 = 0**

**(x ^{2 }+ 8x + 16) - 16 = 0**

**(x + 4) ^{2 }- 16 = 0**

**(x + 4) ^{2} = 16**

** x + 4 = ±4**

**x = −4 ± 4**

So, the solutions to the quadratic equation x^{2 }+ 8x + 16 = 0 are:

**x _{1} = -4 + 4 = 0**

**x _{2} = -4 – 4 = -8**

In this case, there are two distinct solutions:

**x _{1} = 0**

**x _{2} = -8**

** **

** **

**GCSE Past Paper Question 1: Finding the Turning Point**

Question: Find the coordinates of the turning point on the curve with the equation

y = 9 + 18x – 3x^{2}. Show all your working.

Solution:

**Step #1:**Begin with the given equation:

**y = 9 + 18x – 3x ^{2}**

**Step #2:**Express it in standard form by rearranging the terms:

**y = -3x ^{2} + 18x + 9**

**Step #3:**Factor out the leading coefficient ‘a’ (which is -3 in this case) from the first two terms:

**y = -3(x ^{2} – 6x) + 9**

**Step #4:**Complete the square by adding and subtracting (b/2)^{2}inside the parentheses. Here, (6/2)^{2 }= 9:

**y = -3(x ^{2} – 6x + 9 – 9) + 9**

**Step #5:**Rearrange the equation by grouping the terms:

**y = -3((x ^{2} – 6x + 9) – 9) + 9**

**Step #6:**Simplify the grouped terms inside the parentheses:

**y = -3(x – 3) ^{2} – 27 + 9**

**Step #7:**Further simplify:

**y = -3(x – 3) ^{2} – 18**

Now, we have the equation in vertex form, a(x – h)^{2} + k, where a = -3, h = 3,

and k = -18.

The turning point’s coordinates are (h, k) = (3, -18).

So, the coordinates of the turning point on the curve with the equation

y = 9 + 18x – 3x^{2} are (3, -18).

** **

** **

**GCSE Past Paper Question 2: Sketching the Graph**

Question: Sketch the graph of y = 2x^{2 }– 8x – 5, indicating the turning point and providing the precise coordinates of any intercepts with the coordinate axes.

Solution:

To sketch the graph of the given equation, let’s first identify the key elements:

the turning point and the intercepts.

**Turning Point: **

We’ve already found the turning point coordinates in a previous example:

(h, k) = (2, -9).

**Intercepts: **

To find the intercepts, we set y = 0 and solve for ‘x’ (x-intercepts) or set x = 0

and solve for ‘y’ (y-intercepts).

For x-intercepts (when y = 0):

**0 = 2x ^{2 }– 8x – 5**

We can use the quadratic formula to find the roots:

In this case, a=2, b=−8, and c=−5.

Calculating the roots, we get:

Solving these, we find the x-intercepts:

x_{1} ≈ −0.68 (rounded to two decimal places)

x_{2} ≈ 3.18 (rounded to two decimal places)

For y-intercepts (when x = 0):

**y = 2(0) ^{2 }– 8(0) – 5**

**y = -5**

So, the y-intercept is at (0, -5).

Now, we can sketch the graph with these key elements:

- Turning Point: (2, -9)
- X-Intercepts: (-0.68, 0) and (3.18, 0)
- Y-Intercept: (0, -5)

Plot these points and sketch the parabolic curve. The graph of y = 2x^{2 }– 8x – 5 should resemble a parabola opening upward with the specified features.

Question 1: **Turning Point:** Find the coordinates of the turning point on the curve with the equation:

y = 4 - 8x + 3x^{2}

Question 2: **Completing the Square:** Convert the following quadratic equation into vertex form (a(x - h)^{2} + k) by completing the square: y = x^{2} - 6x + 9

Question 3: **Solving Quadratic Equations:** Solve the quadratic equation: x^{2} + 5x - 6 = 0

Question 4: **Graph Sketching:** Sketch the graph of the quadratic equation y = -x^{2} + 4x + 3 and label the coordinates of the turning point and any intercepts with the coordinate axes.

Question 5: **Solving Quadratic Equations: **Solve the quadratic equation: 2x^{2} - 12x + 18 = 0