9709 Pure 1 · 1.1 Quadratics · Session 2 of 3
Khan Academy has no exercise for this skill. Today's hour carries it on its own.
Do Now · 5 min · wkshtH Part B · no notes, no calculator
Question 2 is two conditions joined by and, so both have to hold at once. Question 1 is a single condition. That difference is on purpose.
Do Now · what to keep from it
Direct instruction · 1 of 10
An equation asks where an expression is zero, and the answer is a couple of numbers. An inequality asks where it sits above or below zero, and the answer is a stretch of the number line. Same expression, different question, completely different kind of answer.
A linear expression
| \(x\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| \(2x - 4\) | −4 | −2 | 0 | 2 | 4 |
Negative, then zero, then positive, and it never comes back. One crossing, two regions.
Add 4 to both sides, divide by 2, and \(x\) is sitting on its own: \(x > 2\). The algebra is the answer.
A quadratic
| \(x\) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| \(x^2-5x+4\) | 4 | 0 | −2 | −2 | 0 | 4 |
Positive, zero, negative, zero, positive. It changes its mind. Two crossings, three regions.
No move gets \(x\) on its own. It is squared and not squared in the same line, so there is nothing to divide by that you know the sign of.
So you are not solving for \(x\). You are finding the changeover points, then deciding which regions you want. The rest of the lesson is those two halves.
Direct instruction · 2 of 10
A critical value is a value of \(x\) that makes the expression equal to zero. They are the roots of the matching equation, and they are the only places the sign can change.
Between 1 and 4 the curve is under the axis, so the expression is negative there. Outside them it is above the axis, so it is positive. Nothing changes sign anywhere else, because nowhere else does the curve touch zero.
Direct instruction · 3 of 10
Both routes below need the expression compared to zero, because a critical value is a value that makes it zero. Until one side is 0, there is nothing to find. This first line is worth a mark on its own.
Move the terms across
Turning the symbol around is not a rule you apply. It is the same sentence read backwards. \(3 < 5\) and \(5 > 3\) are one fact, not two, so "0 is greater than or equal to it" and "it is less than or equal to 0" say the same thing.
Why not just divide?
Check \(x = -1\) in the original: \(1 > -4\) is true. Dividing threw that answer away.
Two things went wrong. You do not know the sign of \(x\), so you do not know whether to flip. And dividing by \(x\) deleted a critical value: \(x(x - 4)\) is zero at 0 and 4, while \(x > 4\) has only one boundary left.
Dividing by a number is fine, as long as you flip when it is negative. Dividing by anything with an \(x\) in it is not, ever. And when you move terms across, move them to the side that leaves \(x^2\) positive, so the curve opens upward and the sketch is the easy one.
Direct instruction · 4 of 10 · wkshtJ (a)
Route 1 · sign table on the number line
Critical values: \(x = 1\) and \(x = 4\). They cut the line into three regions. Take the sign of each bracket in each region.
| \(x < 1\) try \(x = 0\) | \(1 < x < 4\) try \(x = 2\) | \(x > 4\) try \(x = 5\) | |
|---|---|---|---|
| \(x - 1\) | −1 | +1 | +4 |
| \(x - 4\) | −4 | −2 | +1 |
| multiply them | +4 | −2 | +4 |
Each bracket flips sign at its own critical value: \(x - 1\) goes positive once \(x\) passes 1, \(x - 4\) not until \(x\) passes 4. Between them one is positive and one is negative, so the product is negative. That is the column we want.
Route 2 · sketch it and read the axis
Same critical values, used as the two places the curve crosses the axis. The \(x^2\) term is \(+1\), so it opens up.
We want the expression below zero, which is the part of the curve below the axis. That is the dip between the two crossings. Read the \(x\) values it sits over.
Strict on both ends. At \(x = 1\) and \(x = 4\) the expression is exactly 0, and 0 is not less than 0, so both crossings are out. That is what the hollow circles mean.
Direct instruction · 5 of 10
You have the critical values. Now you have to pick which regions to keep, and that is two questions, not one. Students lose marks here by guessing "between" or "outside" from memory. You never have to.
Step 1 · the symbol tells you which sign to hunt for
| It says | You want the expression | On the sketch |
|---|---|---|
| \(> 0\) | positive | above the axis |
| \(\ge 0\) | positive, or zero | on or above the axis |
| \(< 0\) | negative | below the axis |
| \(\le 0\) | negative, or zero | on or below the axis |
Read the symbol first, every time. That is the whole of the question: am I hunting for positive or negative?
Step 2 · the curve says where that sign lives
Opens up. Negative is the dip in the middle. Positive is the two arms.
Opens down. Everything swaps: positive is the hump in the middle, negative is the two arms.
So "between or outside" is never a thing to memorize. It falls out of the two steps. The symbol hands you a sign, the curve tells you where that sign lives, and the answer is whatever is left standing.
Do them in that order and you cannot get the between-or-outside call wrong, because you never made the call. You read it.
Write a between answer as one statement: \(1 < x < 4\). Write an outside answer as two, joined by "or": \(x \le 1\) or \(x \ge 4\). It will not fit in one chain: \(4 \le x \le 1\) claims \(x\) is at least 4 and at most 1, which nothing is.
Direct instruction · 6 of 10
Step 1 is already done and it never changes: \(< 0\) means we are hunting for NEGATIVE. The only thing moving is the curve.
Nothing about the question changed. Same critical values, same symbol, same sign being hunted. The answer swapped shape only because the curve flipped over and took the negative part with it. That is step 2 doing its job in front of you.
Direct instruction · 7 of 10 · wkshtJ (b)
Route 1 · sign table
| \(k < 1\) | \(1 < k < 4\) | \(k > 4\) | |
|---|---|---|---|
| \((k-1)(k-4)\) | + | − | + |
Same signs as the last slide, because it is the same expression. Only the letter changed. This time we want the product positive or zero, so we take the two outer columns: \(k < 1\) and \(k > 4\).
Then \(\ge\) lets the critical values in themselves, so each end moves out one notch: \(k \le 1\) and \(k \ge 4\). Two separate pieces of the line, so they are joined by or.
Why does \(k \le 1\) run forever to the left? Check \(k = 0\): \(0 - 0 + 4 = \mathbf{4}\). Check \(k = -10\): \(100 + 50 + 4 = \mathbf{154}\). The left arm climbs and never comes back down.
Route 2 · sketch
Same curve. This time we want it on or above the axis, which is the two arms, and the crossings count because \(\ge\) allows zero. Filled circles mean included.
Perfect factoring gets this one wrong if you copied the shape of the last answer. The factors were identical. Only the direction of the symbol moved, and it moved the answer from the middle to the outside.
Direct instruction · 8 of 10
Both are correct and both earn full marks. They are not equally quick on every question, so here is the rule.
Default: sketch it
Use it when:
About ten seconds, and it is the route Cambridge's own teaching guidance recommends.
Switch to: sign table
Use it when:
It works from the factors alone. It never needs you to picture anything.
The one part that is not a choice sits before either of them: get everything on one side first. Skip that and both routes are wrong from the first line.
Direct instruction · 9 of 10
Read each one and say what your first line would be, and which route you would take. Ten seconds each.
Number 2 is the one that catches people. There is an \(x\) on both sides, and moving the \(6x\) across is the whole first line. Dividing by \(x\) is the trap from earlier. Number 3 is where the Do Now comes back: multiplying by \(-1\) flips the symbol, and it turns a downward parabola into an upward one. It tidies the question. It does not answer it.
Direct instruction · 10 of 10 · back to Session 1
Run the discriminant on the matching equation. A negative discriminant means no real roots, so there are no critical values and the curve never touches the axis.
\(\Delta < 0\), and the \(x^2\) term is \(+1\), so the curve opens up and sits entirely above the axis. The expression has one sign everywhere.
"No solution" is a complete answer here. So is "all real values of \(x\)". Neither one means you got stuck.
Practice · together · 10 min · wkshtJ
(c) is not on one side yet. Expand the bracket, move the 20 across, then start. That first line is worth a mark on its own.
Practice · on your own · 10 min · wkshtJ
(e) is the one to watch. Moving the \(x^2\) across puts the expression on the right, so when you write it on the left the symbol turns around. (f) is the only one of these four that comes out outside the critical values.
Quadratic Inequality Explorer: check your own sketch against it
Exit ticket · 5 min
Both sides of zero count. Check \(x = -11\): \(121 - 100 = 21\), which is greater than 0.
Homework · and where to practice more
Quadratic Inequality Explorer Corbettmaths practice Corbettmaths textbook exercise PMT stretch set
Next session: simultaneous equations, and quadratics wearing a disguise.