9709 Pure 1 · 1.1 Quadratics · Session 1 of 3
Do Now · 5 min · no notes, no calculator
Two of these lose a root if you rush. \(x^2 - 9\) loses the \(-3\) if you only take the positive square root. \(x^2 + 4x\) loses \(x = 0\) if you divide both sides by \(x\). Quadratics come in pairs.
Direct instruction · 1 of 12
Khan Academy drills all three. The skill today is picking the right one.
Direct instruction · 2 of 12
Halve the \(-6\) to get \(-3\). Now check what you built: \((x - 3)^2\) expands to \(x^2 - 6x + \mathbf{9}\). That is 9 more than you wanted, so subtract 9 to put it back. Then combine with the 5.
The \(\equiv\) sign means "is identical to". It is not asking you to solve anything. It says these two are the same expression written two ways, for every value of \(x\).
Squaring anything gives zero or more, never less. So \((x + p)^2\) is at its smallest when it is exactly zero, which happens when \(x + p = 0\), so \(x = -p\). At that \(x\) nothing is left but \(q\). That point \((-p, q)\) is the vertex.
Vertex Form Explorer: drag the vertex and watch the equation change
Direct instruction · 3 of 12
Route 1 · factor the a out
Pull the 3 out of the \(x^2\) and \(x\) terms. Leave the \(-15\) outside.
Half of \(\tfrac{4}{3}\) is \(\tfrac{2}{3}\). Square it, \(\tfrac{4}{9}\), and subtract it inside.
Multiply back out. The \(\tfrac{4}{9}\) sits inside, so it becomes \(\tfrac{4}{3}\).
Add \(\tfrac{49}{3}\) to both sides, then divide by 3.
Route 2 · divide the whole line by a
Divide every term by 3. The other side is 0, and \(\tfrac{0}{3}\) is still 0.
Move the \(-5\) across.
Half of \(\tfrac{4}{3}\) is \(\tfrac{2}{3}\). Square it, \(\tfrac{4}{9}\), and add it to both sides.
Tidy the right side.
Same bracket, same answer. Route 2 is shorter here only because the other side is 0, so dividing costs nothing. On a question like this one, divide.
Direct instruction · 4 of 12
wkshtA and wkshtC never say "= 0". They say write in the form \(A(x + B)^2 + C\). There is no other side to divide into, and setting it equal to zero yourself just solves for \(x\) instead.
Route 1 · factor the a out
Route 2 · divide, then put the a back
Same trap on both routes, and it is the \(a\): Route 1 brings the \(-9\) back out as \(-9\) instead of \(-18\), Route 2 stops one line early. The bracket survives either way, so \(x = 3\) is right. The minimum value is not: it is \(-13\).
So the whole thing comes down to one question, asked before you start. Next slide.
Direct instruction · 5 of 12
You do not choose by looking at the quadratic. You choose by looking at what is being asked. One thing decides it.
If there is an = 0
It looks like one of these:
The other side is 0, so dividing is free. You never need the \(a\) again.
If there is no equals sign
It looks like one of these:
The \(a\) stays on the page the whole way, so there is nothing to put back.
The right-hand list is one question wearing five hats. Vertex, turning point, minimum, maximum, range are all read straight off \(a(x + p)^2 + q\), which is why every one of them needs the \(a\) to be right.
Direct instruction · 6 of 12
Read each one and say which route you would take. That is the whole task. Ten seconds each.
Number 3 is the one that catches people. There is no equals sign anywhere in it, and the minimum value is the \(q\), so you need the \(a\) on the page to get it right.
Direct instruction · 7 of 12
The vertex does not tell you which way it opens. The sign of \(a\) does that.
To sketch any quadratic
Which way it opens (sign of \(a\)) · the vertex · the \(y\)-intercept (put \(x = 0\), that is \(c\)).
Opens up, \(a = 2\). Vertex \((3, -13)\). Crosses the \(y\)-axis at 5. Three facts and you have the curve.
Direct instruction · 8 of 12
Physics · projectile motion
A ball's height in meters, \(t\) seconds after launch:
\(a\) is negative, so the vertex is the maximum. Highest point 21 m, reached at \(t = 2\) s.
Business · maximizing profit
Daily profit in dollars from selling \(x\) units:
Maximum profit $3500, at 30 units. Sell more than that and profit falls again.
Both have \(a < 0\), so both open downward and both vertices are maximums. That has to be true: a ball comes back down, and profit falls once you overproduce. If a question asks for a maximum and your \(a\) comes out positive, you have made a sign error. Both answers came straight out of the completed square, with no calculus.
Direct instruction · 9 of 12
A root is a value of \(x\) that makes the expression equal zero. On the graph, the roots are where the curve crosses the \(x\)-axis, because the \(x\)-axis is the line \(y = 0\).
The \(\pm\) is what gives you two answers. \(\Delta\) decides whether it can.
The \(-b\) and the \(2a\) move the answers around. Only \(\Delta\) changes how many there are.
Direct instruction · 10 of 12
Say "repeated root". It is the syllabus's own term and the exam expects it.
Direct instruction · 11 of 12
A meeting point sits on both graphs at the same time. So it has one \(y\)-value, and both equations have to give it.
On the line, its \(y\) is 4. On the curve, its \(y\) is \(x^2\). Same point, same \(y\).
That is what "set line = curve" means. Take the right-hand side of each equation and put an equals sign between them. The \(y\) drops out because it was the same on both.
Solve it: \(x = 2\) or \(x = -2\). They meet at \((-2, 4)\) and \((2, 4)\).
The exam version is the same two steps. \(y = kx - k\) and \(y = -\tfrac{1}{2x}\) both start with \(y\), so \(kx - k = -\tfrac{1}{2x}\).
Direct instruction · 12 of 12
The \(x\)-axis is just the line \(y = 0\). It is not special. Swap it for any other line and everything carries over.
Same curve, \(y = x^2\). Three different lines. Make the right-hand sides equal each time, then count the solutions.
In all three, \(a = 1\) and \(b = 0\), so \(\Delta = b^2 - 4ac\) collapses to just \(-4c\).
Direct instruction · 12 of 12 · continued
Making the two sides equal builds a new equation. Its roots are the meeting points, and \(\Delta\) counts them.
The classic error: taking \(\Delta\) of the curve alone. That answers "where does the curve cross the \(x\)-axis". Your line never entered the calculation, so it scores nothing.
Direct instruction · outside class · engineering
A dish is a parabola. A signal arriving straight on bounces off toward one focus point. To know the angle it bounces at, you need the tangent line at the spot where it hits.
Tangent line: \(y = x - 1\). Check it: \(\tfrac{x^2}{4} = x - 1\) gives \((x - 2)^2 = 0\), one repeated root at \(x = 2\), exactly the point we started from.
Direct instruction · outside class · business
Revenue from selling \(x\) units is \(R(x) = -5x^2 + 300x\). Fixed cost is \(C\). The business breaks even when revenue exactly equals cost, so profit is zero.
The vertex says the same thing. Completing the square gives \(P = -5(x - 30)^2 + (4500 - C)\), and that vertex height hits zero at exactly \(C = 4500\).
Worked example · 10 min · real past paper
The line with equation \(y = kx - k\), where \(k\) is a positive constant, is a tangent to the curve with equation \(y = -\dfrac{1}{2x}\).
Find, in either order, the value of \(k\) and the coordinates of the point where the tangent meets the curve.
Before any algebra: what does "tangent" translate to? The two sides, made equal, must have exactly one solution.
Worked example · step by step
\(a = 2k\) · \(b = -2k\) · \(c = 1\). \(k\) is the unknown you are solving for and part of \(a\) and \(b\).
Both are real answers to that equation. The question said \(k\) is a positive constant, so \(k = 0\) is rejected. Always check your answers against the wording.
\((2x - 1)^2 = 0\) has one repeated root, which is the algebra confirming it touches once. Mark scheme accepts a quadratic in \(x\) or in \(y\).
Practice · together · 10 min · wkshtA Part 3, Q3–4
Numbers 3 and 4 say "write in the form", so factor the \(A\) out. Number 5 has an equals sign, so divide. It is also wkshtC number 4 with an equals sign bolted on. Same quadratic, different question, different route.
Number 4 gives you a fraction inside the bracket. Keep it as a fraction. Halving \(-3\) gives \(-\tfrac{3}{2}\), not \(-1.5\) and not \(-1\).
Practice · on your own · 10 min · wkshtC Q4–8
No equals signs on this sheet, so whichever route you take, the number in front has to come back.
Then this one
"Values" is plural. If you found one \(k\), you're halfway.
Stop on number 7 when you get there. \(a\) is positive and so is the number outside the bracket, so the whole thing can never be zero. Then work out \(\Delta\) and watch it tell you the same thing.
Extra practice · if time, or take it home
Number 3 has \(a \ne 1\), so the two numbers you need multiply to \(a \times c = -45\) and add to 4. It is also the one we completed the square on both ways earlier. Three routes, one answer. Factoring wins when it factors, and this one does.
Exit ticket · 5 min
Homework: finish wkshtC. Extension: Underground Mathematics "Discriminating" card sort.
Vertex Form Explorer Quadratics Sprint Two Routes, One Answer One-page summary (print it)