Understand · explore · practise
Rewrite a quadratic to uncover its turning point, solve equations and explain why the method works.
Before you startExpanding brackets
Squares and square roots
01 / The idea
A quadratic can look complicated even when it is just a squared bracket with a number added or subtracted.
x² + 6x + 4 = (x + 3)² − 5
These expressions have the same value for every x. The second form tells us something new: a square cannot be negative, so the expression can never fall below −5.
Why is there a 3 in the bracket? Expanding (x + 3)² produces two lots of 3x. Together they give the 6x we need.
The area picture uses positive lengths. The algebraic identity remains true for negative x as well.
The extra 9 completes the square. Subtract it again to keep the original expression unchanged.
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02 / The method
Complete the square for x² + 6x + 4. Keep each line equal to the one before it.
Halve the coefficient of x.
Half of 6 is 3, so try (x + 3)².
(x + 3)² = x² + 6x + 9
The squared bracket brings an extra 9.
x² + 6x + 4 = (x + 3)² − 9 + 4
Remove that 9, then keep the original +4.
= (x + 3)² − 5
Combine the constants. Nothing else changes.
Check by expanding: x² + 6x + 9 − 5 gives the original x² + 6x + 4.
The extra 9 completes the square. Subtract it again to keep the original expression unchanged.
Why not (x + 6)²?
It expands to x² + 12x + 36. The middle term would be twice as large as we need.
03 / Other coefficients
When the coefficient of x² is not 1, factor it out of the x² and x terms before completing the square inside the bracket.
2x² − 12x + 7 = 2(x² − 6x) + 7
The number to halve is now −6. It gives −3, so the squared bracket is (x − 3)².
Subtract 9 inside the bracket. Since the whole bracket is multiplied by 2, that subtraction becomes −18 outside.
A common slip: subtracting only 9 outside would change the original expression.
Use fractions rather than rounding. For example:
x² + 5x + 2 = (x + 5/2)² − 17/4
Half of 5 is 5/2; its square is 25/4. The constant is 2 − 25/4 = −17/4.
2x² − 12x + 7
Factor out 2 from the terms containing x.
= 2(x² − 6x) + 7
Complete the square inside the bracket.
= 2[(x − 3)² − 9] + 7
Multiply both terms in the bracket by 2.
= 2(x − 3)² − 18 + 7
= 2(x − 3)² − 11
The multiplier stays in front of the squared bracket.
04 / Explore the graph
In y = a(x − h)² + k, the square becomes zero at x = h. That puts the turning point at (h, k).
y = (x + 3)² − 5
Turning point: (−3, −5)
With a positive multiplier, the graph opens upwards and k is its minimum. With a negative multiplier, it opens downwards and k is its maximum.
Try this: change a to −1. Keep the turning point in the same place. What changes about the possible values of y?
Move the sliders or drag the turning point. You can use the arrow keys on either slider. The dashed line is the axis of symmetry.
For x² + 6x + 4 = 0:
(x + 3)² = 5
x + 3 = ±√5
x = −3 ±√5
Both signs matter: there are two crossings of the x-axis. If a square must equal a negative number, there are no real solutions.
y = (x + 3)² − 5
The minimum is −5, reached when x = −3.
05 / Your turn
Work on paper. Use a hint if you need a starting point, then compare each line with the solution.
Check a completed-square answer by expanding it. Check a turning point by making the squared bracket equal zero.
The questions move from a direct calculation to an error check and a small design problem.
Write in completed-square form, then state the minimum value and where it occurs.
x² − 10x + 31
Half of −10 is −5. What constant appears when you expand (x − 5)²?
x² − 10x + 31
= (x − 5)² − 25 + 31
= (x − 5)² + 6
The minimum is 6, when x = 5. The squared term is never negative and is zero at that value of x.
A student writes 3x² + 12x + 1 = 3(x + 2)² − 3. Find the error and correct the answer.
Expand the proposed answer first. Compare its constant term with the original.
3(x + 2)² − 3 = 3x² + 12x + 9
The constant is 9, not 1. Completing the square correctly gives:
3(x² + 4x) + 1
= 3[(x + 2)² − 4] + 1
= 3(x + 2)² − 11
The subtraction of 4 happens inside a bracket multiplied by 3.
A quadratic has coefficient −2 in front of x² and a maximum of 9 at x = 4. Find its expanded form and its two real roots.
Start with the turning-point form a(x − h)² + k. The maximum tells you h and k.
y = −2(x − 4)² + 9
= −2x² + 16x − 23
For the roots, set y = 0:
(x − 4)² = 9/2
x = 4 ± 3√2/2
The negative coefficient makes the turning point a maximum. Both roots lie the same distance from x = 4.
Find the minimum of x² − 7x + 15. Use fractions, not rounded decimals.
Half of −7 is −7/2. Square that fraction carefully.
x² − 7x + 15
= (x − 7/2)² − 49/4 + 60/4
= (x − 7/2)² + 11/4
The minimum is 11/4 at x = 7/2. It is positive, so the equation x² − 7x + 15 = 0 has no real solutions.
06 / What to remember
ax² + bx + c = a(x + b/(2a))²
+ c − b²/(4a), a ≠ 0
You can derive this general form using the same steps. Understanding the correction is more useful than remembering the formula on its own.
Section 1 of 6 · The idea