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A short class with Sarah and her students. The other ways are under More.
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Try to factorise x² + 4x + 1. Which numbers multiply to 1 and add to 4?
Only 1 and 1 multiply to 1, and they add to 2. Nothing works.
Right, the solutions are not whole numbers. So we will build a perfect square instead.
What makes x² + 4x into a perfect square?
Remember (x + 2)² = x² + 4x + 4. So x² + 4x is just 4 short of a perfect square.
So we add 4, which is half of 4, squared.
Move the constant over first. What do we get?
x² + 4x = −1. Then add 4 to both sides: x² + 4x + 4 = 3.
Write the left side as a square.
(x + 2)² = 3. Then x + 2 = √3.
Careful: what else squares to give 3?
−√3 as well! So x + 2 = ±√3 and x = −2 ± √3.
That is about −0.268 and −3.73 as decimals.
We showed x² + 4x + 1 = (x + 2)² − 3. What is the smallest value this can take?
(x + 2)² is at least 0, so the smallest value is −3, when x = −2.
So the vertex is (−2, −3). Now complete the square for x² + 2x + 5.
(x + 1)² + 4. Setting it to zero gives (x + 1)² = −4.
Can a square be negative?
No, so there are no real solutions. The vertex (−1, 4) is above the x-axis.
to complete x² + bx, add (b/2)², because (x + b/2)² = x² + bx + (b/2)².
whatever is added to one side of an equation must be added to the other.
taking the square root of both sides gives two solutions, so write ±.
if the coefficient of x² is not 1, divide every term by it before completing the square.
y = (x + h)² + k has vertex (−h, k), so completing the square also locates the turning point.
if completing the square leads to a square equal to a negative number, the quadratic has no real roots.
Solve x² + 4x + 1 = 0: x² + 4x + 4 = 3, so (x + 2)² = 3 and x = −2 ± √3 (about −0.268 and −3.73).
Solve x² − 6x + 2 = 0: (x − 3)² − 9 + 2 = 0, so (x − 3)² = 7 and x = 3 ± √7 (about 5.65 and 0.354).
Solve 2x² − 8x + 3 = 0: divide by 2 to get x² − 4x + 3/2 = 0, then (x − 2)² = 5/2, so x = 2 ± √(5/2) (about 3.58 and 0.419).
Show x² + 2x + 5 = 0 has no real solutions: (x + 1)² + 4 = 0 means (x + 1)² = −4, which is impossible, so there are none.
Halve it, square it, add it in; balance both sides, then root it with a grin; plus or minus, don't forget, or half your answers you won't get.
A gardener has a square bed x by x and two strips 2 by x along two sides. One 2 by 2 corner is bare. She plants 4 more square metres, and suddenly the whole bed is a perfect square of side x + 2.
"Move, Halve, Square, Add, Root" lists the steps in order: move the constant, halve b, square it, add to both sides, take ± square roots.
H.S.A. - Halve, Square, Add: the three moves that complete the square.
Term
What number completes the square for x² + bx?
Click to flip
Term
Write x² + 4x + 1 in completed-square form.
Click to flip
Term
Solve (x + 2)² = 3.
Click to flip
Term
What is the vertex of y = (x + 2)² − 3?
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Term
What does it mean if completing the square gives (x + 1)² = −4?
Click to flip
Rewriting x² + bx + c as (x + b/2)² + k so that a quadratic can be solved by square roots.
An expression that equals a bracket squared, such as x² + 4x + 4 = (x + 2)².
An irrational root left in exact form, such as √3.
An answer written with surds or fractions rather than rounded decimals, such as −2 ± √3.
The turning point of a parabola; for y = (x + h)² + k it is (−h, k).
The vertical line through the vertex, about which the parabola is symmetric.
A solution that is a real number; a quadratic has none when a square would have to be negative.
Example 2 solves x² + 4x + 1 = 0 by completing the square: the equation becomes (x + 2)² = 3, so x = −2 ± √3. Completing the square solves quadratics that do not factorise and also gives the vertex of the parabola.
Some quadratics, like x² + 4x + 1 = 0, cannot be factorised with whole numbers because their solutions are irrational. Completing the square rewrites the left side as a perfect square plus a number: x² + 4x + 1 = (x + 2)² − 3. Setting this to zero gives (x + 2)² = 3, and taking square roots gives the exact solutions x = −2 + √3 and x = −2 − √3, about −0.268 and −3.73. This lesson works through the example step by step, explains why we add the square of half the x-coefficient, and applies the method to odd coefficients, a leading coefficient other than 1, an equation with no real solutions, and finding the vertex of a parabola.
This lesson is the second of three worked examples on solving quadratic equations. Example 1 used factoring; here we use completing the square, and Example 3 uses the quadratic formula.
The idea. A perfect square expands in a fixed pattern: (x + h)² = x² + 2hx + h². So in x² + 4x, the 4x must be 2hx with h = 2, and the square is completed by adding h² = 4. In general, for x² + bx we add (b/2)², the square of half the coefficient of x.
Example 2: Solve x² + 4x + 1 = 0.
As decimals, x ≈ −0.268 or x ≈ −3.73 (3 significant figures). Exact answers in surd form are usually preferred unless a question asks for decimals.
Checking. For x = −2 + √3: x² = 4 − 4√3 + 3 = 7 − 4√3 and 4x = −8 + 4√3, so x² + 4x + 1 = 7 − 4√3 − 8 + 4√3 + 1 = 0.
A bonus: the vertex. The same working shows that y = x² + 4x + 1 = (x + 2)² − 3. Since (x + 2)² is never negative, the smallest value of y is −3, reached when x = −2. So the vertex is (−2, −3) and the axis of symmetry is x = −2. The two solutions sit symmetrically either side of x = −2, at a distance of √3.
More worked examples.
Common mistakes. Forgetting the ± when taking square roots (which loses one solution), adding (b/2)² to one side only, and completing the square before dividing through by a leading coefficient that is not 1.