Algebra
Grade 7-8Completing the Square
Completing the square rewrites a quadratic expression as a single bracket squared plus a constant, revealing the minimum point of its graph directly and giving another way to solve a quadratic equation. This lesson covers completing the square when the coefficient of x² is 1, finding the minimum point of a quadratic graph from the completed square form, completing the square when the coefficient of x² is greater than 1, and solving a quadratic equation by completing the square.
Written by Asad, Co-Founder of Teachably
Video walkthrough coming soon
The written lesson below covers everything you need in the meantime.
Completing the square when the coefficient of x² is 1
Halve the coefficient of x to find the value inside the bracket, square this bracket, then subtract the square of that value and add back the original constant.
To write x² + 8x + 5 in the form (x + a)² + b, halve 8 to get 4, so the bracket is (x + 4)². Since (x + 4)² = x² + 8x + 16, subtract 16 and add the original 5: x² + 8x + 5 = (x + 4)² − 11.
Finding the minimum point from the completed square form
Once a quadratic is written as (x + a)² + b, the minimum point of its graph is at (−a, b), since the squared bracket is smallest, equal to 0, when x = −a.
For y = x² + 8x + 5 = (x + 4)² − 11, the minimum point is at (−4, −11).
Completing the square when the coefficient of x² is greater than 1
Factor the coefficient of x² out of the x² and x terms first, complete the square inside the bracket, then multiply back through and simplify.
To write 2x² + 8x + 3 in the form p(x + q)² + r, factor out 2: 2(x² + 4x) + 3. Completing the square inside gives 2[(x + 2)² − 4] + 3, which simplifies to 2(x + 2)² − 5.
Solving a quadratic equation by completing the square
Complete the square, set the result equal to zero, then rearrange to isolate x, taking the square root of both sides and remembering both the positive and negative root.
To solve x² + 6x + 2 = 0, complete the square: (x + 3)² − 7 = 0, so (x + 3)² = 7. Taking the square root of both sides gives x + 3 = ±√7, so x = −3 + √7 or x = −3 − √7.
Worked Examples
Three exam-style questions, fully solved.
x² + 8x + 5 can be written in the form (x + a)² + b, where a and b are integers. Find the values of a and b.
Easy- 1.Halve the coefficient of x: 8 ÷ 2 = 4
- 2.Write the bracket and subtract its square, adding back the constant: (x + 4)² − 16 + 5
Answer: a = 4, b = −11
2x² − 12x + 5 can be written in the form p(x + q)² + r, where p, q and r are integers. Find the values of p, q and r.
Medium- 1.Factor out the coefficient of x²: 2(x² − 6x) + 5
- 2.Complete the square inside the bracket: 2[(x − 3)² − 9] + 5
- 3.Multiply back through and simplify: 2(x − 3)² − 18 + 5 = 2(x − 3)² − 13
Answer: p = 2, q = −3, r = −13
Write x² + 6x + 2 in the form (x + a)² + b, and hence solve x² + 6x + 2 = 0, giving your answers in surd form.
Hard- 1.Complete the square: (x + 3)² − 9 + 2 = (x + 3)² − 7
- 2.Set the equation to zero and rearrange: (x + 3)² = 7
- 3.Take the square root of both sides: x + 3 = ±√7
Answer: x = −3 + √7 or x = −3 − √7
Avoid These
The most common mistakes students make.
Getting the sign inside the bracket wrong, such as writing (x + 4)² for x² − 8x + ... instead of (x − 4)².
Forgetting to subtract the square of the halved value after writing the bracket, leaving (x + 4)² instead of (x + 4)² − 16 for x² + 8x.
When the coefficient of x² is greater than 1, forgetting to factor it out of the x² and x terms before completing the square, or making an error multiplying back through.
Misreading the coordinates of the minimum point from the completed square form, writing (a, b) instead of (−a, b).
When solving using the completed square form, forgetting to include both the positive and negative square root, or making an arithmetic slip isolating x.
FAQ
Questions parents and students ask.
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