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Completing the Square – Sec 3 G2 and G3 Question Explained

Source: Completing the Square Sec 3 G2 and G3

Introduction

This completing the square question is an important Secondary Maths algebra skill because students must rewrite a quadratic expression into a new form without changing its value. Many students understand the idea, but lose marks when they add the correct square term and forget to subtract it back. Once the method is written carefully, this completing the square question becomes much more systematic.

 

completing the square Sec 3 G2 and G3 explained

 

The Question / Scenario Explanation

Source: Completing the Square Sec 3 G2 and G3

Question (as shown):
(a) Express the following in the form \((x+a)^2 + b\).
(i) \(x^2 – 9x – 6\)

 

Step-by-Step Solution / Explanation

Step 1: Identify half of the coefficient of \(x\)

For this completing the square question, look at the coefficient of \(x\), which is \(-9\).

Half of \(-9\) is:

\(-\frac{9}{2}\)

This tells us the square bracket will be:

\(\left(x-\frac{9}{2}\right)^2\)

Step 2: Add and subtract the square term

Start with:

\(x^2 – 9x – 6\)

Add and subtract \(\left(\frac{9}{2}\right)^2\):

\(x^2 – 9x + \left(\frac{9}{2}\right)^2 – \left(\frac{9}{2}\right)^2 – 6\)

This keeps the expression equal while allowing us to form a perfect square.

Step 3: Write the perfect square bracket

The first three terms become:

\(\left(x-\frac{9}{2}\right)^2\)

So now we have:

\(\left(x-\frac{9}{2}\right)^2 – \frac{81}{4} – 6\)

Step 4: Simplify the constant part

Convert \(6\) into quarters:

\(6 = \frac{24}{4}\)

So:

\(-\frac{81}{4} – 6 = -\frac{81}{4} – \frac{24}{4} = -\frac{105}{4}\)

Therefore:

\(x^2 – 9x – 6 = \left(x-\frac{9}{2}\right)^2 – \frac{105}{4}\)

✅ Final Answer: \(\left(x-\frac{9}{2}\right)^2 – \frac{105}{4}\)

Step 5: Match it to the form \((x+a)^2 + b\)

The required form is \((x+a)^2 + b\).

From \(\left(x-\frac{9}{2}\right)^2 – \frac{105}{4}\), we can see:

\(a = -\frac{9}{2}\)

\(b = -\frac{105}{4}\)

This is why the expression fits the required completing the square form correctly.

 

Key Concepts Students Must Know

  • In a completing the square question, half the coefficient of \(x\) is the key step.
  • The general pattern is \(x^2 + px = \left(x+\frac{p}{2}\right)^2 – \left(\frac{p}{2}\right)^2\).
  • After forming the square bracket, always simplify the remaining constants carefully.
  • The final form \((x+a)^2 + b\) may include negative values for both \(a\) and \(b\).
  • This method is useful for graph work, quadratic equations, and algebraic transformation.

 

Exam Tips / Common Mistakes

Exam Tips

  • Take half of the \(x\)-coefficient first before doing anything else.
  • For this completing the square question, half of \(-9\) is \(-\frac{9}{2}\), not \(-9\).
  • Write the add-and-subtract step clearly to show the expression stays equal.
  • Change whole numbers into fractions carefully when combining constants.
  • Check that your final answer is exactly in the form \((x+a)^2 + b\).

Common Mistakes

  • Using \(\left(\frac{9}{2}\right)\) instead of \(\left(\frac{9}{2}\right)^2\).
  • Writing \(\left(x-\frac{9}{2}\right)^2\) but forgetting to subtract \(\left(\frac{9}{2}\right)^2\) as well.
  • Making sign mistakes when simplifying \(-\frac{81}{4} – 6\).
  • Giving \(\left(x-\frac{9}{2}\right)^2 – \frac{81}{4}\) and forgetting the extra \(-6\).

 

Parent Insight

This completing the square question shows why strong algebra habits matter in Secondary Maths. Many students can follow an example, but they lose marks when they rush the fraction step or forget one part of the balancing process. With regular guided practice, students become more confident in rewriting quadratics accurately and recognising standard algebra patterns.

 

Conclusion

To solve this completing the square question, we took half of \(-9\), squared it, and added and subtracted the same value. That allowed us to rewrite \(x^2 – 9x – 6\) as \(\left(x-\frac{9}{2}\right)^2 – \frac{105}{4}\). So the expression in the form \((x+a)^2 + b\) is \(\left(x-\frac{9}{2}\right)^2 – \frac{105}{4}\).

 

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Frequently Asked Questions

Because the square bracket \((x+a)^2\) expands to \(x^2 + 2ax + a^2\). So the coefficient of \(x\) comes from \(2a\), which is why we take half of the \(x\)-coefficient first.

Because the original expression also has a \(-6\). After forming the square bracket, we still need to include that \(-6\), which changes the constant part to \(-\frac{105}{4}\).

Check that there is one square bracket and one constant outside it. Here, \(\left(x-\frac{9}{2}\right)^2 – \frac{105}{4}\) matches \((x+a)^2 + b\) with \(a=-\frac{9}{2}\) and \(b=-\frac{105}{4}\).