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Complete the square to reach vertex form and read off maxima and minima, then solve systems of a line with a parabola or a circle and quadratic inequalities.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you take a quadratic apart and put it back together in the form that answers questions. Completing the square hides every $x$ inside one squared bracket, and once it is there the vertex, the axis of symmetry and the greatest or least value can be read straight off — which is how a projectile's highest point or a cost's minimum is found without any guessing. Then you point the same algebra at systems: substituting a line into a parabola or a circle leaves a single quadratic, whose roots are the meeting points and whose discriminant says how many there are, and the same factoring solves quadratic inequalities by asking where the parabola sits below the axis.
You can expand $(x + 5)^2$ to $x^2 + 10x + 25$ and see that the middle term is twice the $5$. You can solve $x^2 = 49$ by taking both square roots. You know a parabola has a lowest or highest point. Completing the square is those three facts run backwards: given the first two terms of an expansion, rebuild the bracket they came from, then read the turning point straight off.
Perfect square trinomial: a quadratic that is exactly a squared bracket, like $x^2 + 10x + 25$.
Completing the square: rewriting $ax^2 + bx + c$ as $a(x - h)^2 + k$.
Vertex: the turning point of the parabola, at $(h, k)$.
Vertex form: $y = a(x - h)^2 + k$, which shows the vertex without any work.
Axis of symmetry: the vertical line $x = h$ through the vertex.
Maximum / minimum: the value $k$, a maximum when $a < 0$ and a minimum when $a > 0$.
Any quadratic can be written as one squared bracket plus a number.
Monic case. In $x^2 + bx + c$, half of $b$ goes inside the bracket. $(x + \frac{b}{2})^2$ has the right $x^2$ and $x$ terms but an extra $\frac{b^2}{4}$, so subtract it:
$$x^2 + bx + c = \left(x + \tfrac{b}{2}\right)^2 + c - \tfrac{b^2}{4}.$$
So $x^2 + 8x + 3 = (x + 4)^2 + 3 - 16 = (x + 4)^2 - 13$.
With a leading coefficient. Factor $a$ out of the first two terms, complete the square inside, then multiply back out: $3x^2 - 12x + 5 = 3(x^2 - 4x) + 5 = 3\big((x - 2)^2 - 4\big) + 5 = 3(x - 2)^2 - 7$.
Why bother. In $y = a(x - h)^2 + k$ the squared bracket is never negative, so $a(x - h)^2$ never changes sign. If $a > 0$ the smallest $y$ can be is $k$, reached when $x = h$; if $a < 0$ that same $k$ is the largest. The vertex is $(h, k)$ and the axis of symmetry is $x = h$ — no calculus, no table of values. The form also solves the equation directly: $2(x - 3)^2 - 8 = 0$ gives $(x - 3)^2 = 4$, so $x = 3 \pm 2$.
Another way: picture
An $x$ by $x$ square with two strips of width $\frac{b}{2}$ laid along two of its sides. The three pieces make an $(x + \frac{b}{2})$ square except for one missing corner of area $\frac{b^2}{4}$ — which is exactly the amount the algebra has to subtract again.
Another way: story
The bracket hides the variable in one place. Once $x$ appears only inside $(x - h)^2$, its effect on $y$ is a single non-negative number, and the best you can do is make it zero.
Adding the square and forgetting to take it back. $x^2 + 6x + 1$ is not $(x + 3)^2 + 1$. The bracket smuggles in $9$, so you must subtract it: $(x + 3)^2 - 8$. Check by expanding, every time.
Halving $b$ before dividing by $a$. For $2x^2 + 12x + 5$ the number in the bracket is not $6$. Factor the $2$ out of the first two terms first: $2(x^2 + 6x) + 5 = 2(x + 3)^2 - 13$.
Reading $h$ with the wrong sign. The form is $a(x - h)^2 + k$, so $y = (x + 4)^2 - 1$ has $h = -4$ and vertex $(-4, -1)$. The number you see in the bracket is $-h$, not $h$.
Half of $-10$ is $-5$, so the bracket is $(x - 5)$.
The sign of the $x$-term carries into the bracket.
$(x - 5)^2 = x^2 - 10x + 25$, which is $25$ instead of $7$.
Expand to see what the bracket brought in.
So $x^2 - 10x + 7 = (x - 5)^2 + 7 - 25 = (x - 5)^2 - 18$. Vertex $(5, -18)$.
Factor $-2$ from the first two terms: $-2(x^2 - 4x) + 1$.
Dividing $8$ by $-2$ gives $-4$, not $4$.
Inside, $x^2 - 4x = (x - 2)^2 - 4$, so $y = -2\big((x - 2)^2 - 4\big) + 1 = -2(x - 2)^2 + 9$.
Multiply the $-4$ back by $-2$ to get $+8$.
The squared term is never positive, so the greatest $y$ is $9$, at $x = 2$.
Half of $6$ is $3$, and $(x + 3)^2 = x^2 + 6x + 9$.
So $y = (x + 3)^2 + 2 - 9 = (x + 3)^2 - 7$: vertex $(-3, -7)$, a minimum.
Complete the square: fill in the missing number.
x^2 + 2x - 4 = (x + 1)^2 + (k)
$y = 4x^2 + 24x + 28$ can be written in the vertex form $y = 4(x - h)^2 + k$. Find $h$ and $k$.
h = vh, k = vk
A ball's height above the ground $t$ seconds after it is thrown is $h = -3t^2 + 6t + 33$ metres. What is its greatest height?
The greatest height is answer metres.
Complete the square: fill in the missing number.
x^2 + 10x + 1 = (x + 5)^2 + (k)
You solve a pair of straight lines by substitution: rearrange one equation and put it into the other. You solve quadratics by factoring, by the formula, or by completing the square, and you know the discriminant $b^2 - 4ac$ says how many real roots there are. You know $x^2 + y^2 = r^2$ is a circle of radius $r$ about the origin. Nothing new is needed here — only the same substitution, aimed at a curve.
Nonlinear system: two equations, at least one of them not a straight line.
Solution: a pair $(x, y)$ satisfying both — a point where the graphs meet.
Substitution: replacing $y$ in one equation by what the other says it equals.
Discriminant: $b^2 - 4ac$; positive gives two meeting points, zero gives one (a tangent), negative gives none.
Quadratic inequality: $ax^2 + bx + c < 0$ or $> 0$; the answer is a set of $x$, written as an interval.
Substitution still works. With $y = x^2$ and $y = 4x - 3$, the two right-hand sides are both $y$, so they are equal: $x^2 = 4x - 3$. Move everything to one side, $x^2 - 4x + 3 = 0$, factor, and $x = 1$ or $x = 3$. Then go back to the simpler equation for each $y$: the meeting points are $(1, 1)$ and $(3, 9)$.
A circle behaves the same way. For $x^2 + y^2 = 20$ and $y = 2x$, replacing $y$ gives $x^2 + 4x^2 = 20$, so $5x^2 = 20$, $x^2 = 4$ and $x = \pm 2$, with $y = \pm 4$: the points $(2, 4)$ and $(-2, -4)$.
Counting without solving. The substitution always leaves one quadratic, and its discriminant answers "how many?" without any roots being found. Negative means the line misses the curve entirely; zero means it just touches.
Inequalities. To solve $x^2 - x - 6 < 0$, factor: $(x + 2)(x - 3) < 0$, with zeros $-2$ and $3$. The parabola opens upwards, so it is below the axis between the zeros and above it outside them. The answer is $-2 < x < 3$. Sketching the parabola — up or down, and where it cuts — decides every such question.
Another way: picture
A parabola with the line $y = 4x - 3$ drawn across it, cutting at $(1, 1)$ and $(3, 9)$; below, the same parabola shifted so a second line slides down to touch at one point and then falls clear of it — two roots, one root, none.
Another way: story
Substitution is a way of asking the two graphs the same question at once. Whatever is left is a single equation in $x$, and its roots are the shadows of the meeting points on the $x$-axis.
Expecting one answer. A line can cut a parabola or a circle twice. Finding one $x$ and stopping loses half the answer.
Forgetting $y$. A solution is a point. Having found $x = 3$, put it back into the simpler equation to get $y$; the answer is $(3, y)$, not $3$.
Testing signs instead of thinking about the parabola. For $(x - 2)(x - 5) < 0$ the roots are $2$ and $5$, and an upward parabola is below the axis between its roots: $2 < x < 5$. Do not write $x < 2$ or $x > 5$ — that is where it is above. If the inequality were $> 0$, those two outside pieces would be the answer.
Set them equal: $x^2 + 1 = 3x - 1$, so $x^2 - 3x + 2 = 0$.
Everything to one side before factoring.
$(x - 1)(x - 2) = 0$, so $x = 1$ or $x = 2$.
Put each back into $y = 3x - 1$: the points are $(1, 2)$ and $(2, 5)$.
Use the easier equation, and check in the other.
Substitute: $x^2 + (x + 4)^2 = 4$, so $2x^2 + 8x + 12 = 0$, or $x^2 + 4x + 6 = 0$.
Expand the bracket and collect.
Discriminant $16 - 24 = -8$, which is negative.
No need to solve anything.
No real solutions: the line passes the circle without touching it.
Factor: $(x + 2)(x - 4) \geq 0$, with zeros $-2$ and $4$.
An upward parabola is above or on the axis outside its roots: $x \leq -2$ or $x \geq 4$.
Solve the system $y = x^2$ and $y = -8x - 15$. Give the two $x$-values, smaller first.
x = w1 and x = w2
Solve the system $x^2 + y^2 = 40$ and $y = 3x$, giving the solution with $x > 0$.
x = cx, y = cy
Solve the inequality $x^2 + x - 30 < 0$. Fill in the two ends of the interval.
lo < x < hi
Solve the system $y = x^2$ and $y = -5x + 6$. Give the two $x$-values, smaller first.
x = w1 and x = w2
On a flight planner's map the base sits at the origin and a circular no-fly zone around it is the disc $x^2 + y^2 \leq 94$. A straight supply route runs along the line $y = 3x + 10$. Does the route enter the zone?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$y = 3x^2 - 24x + 48$ can be written in the vertex form $y = 3(x - h)^2 + k$. Find $h$ and $k$.
h = vh, k = vk
Solve the system $x^2 + y^2 = 68$ and $y = 4x$, giving the solution with $x > 0$.
x = cx, y = cy
You can complete the square and solve a system with a curve in it. Put $y = x^2 - 6x + 5$ into vertex form and give its minimum, and find where $y = 2x + 3$ meets $y = x^2$.
8. Your turn: put $y = x^2 + 6x + 2$ in vertex form, step 2
19. Your turn: solve $x^2 - 2x - 8 \geq 0$, step 2