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A short class with Sarah and her students. The other ways are under More.
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We want x² − 5x + 6 < 0. Where would be a good place to start?
Solve x² − 5x + 6 = 0 first. It factorises as (x − 2)(x − 3), so x = 2 or x = 3.
Those are the critical values. Why are they so important?
That's where the expression is zero, so it's the only place it can switch from positive to negative.
So the number line splits into x < 2, between 2 and 3, and x > 3.
And in each piece the sign stays the same, so one test value per piece is enough.
Test x = 1, x = 2.5 and x = 4.
At 1 we get (−1)(−2) = 2. At 2.5 we get (0.5)(−0.5) = −0.25. At 4 we get (2)(1) = 2.
Which interval gives a negative value?
Only the middle one, between 2 and 3.
Do we include 2 and 3 themselves?
No. At 2 and 3 the expression is 0, and 0 is not less than 0. So the answer is 2 < x < 3.
Sketch y = x² − 5x + 6. Which way does it open?
Upwards, because the x² term is positive. It crosses the x-axis at 2 and 3.
So where is the graph below the axis?
Between the roots. That matches 2 < x < 3.
What if the question had said greater than zero?
Then we want where it's above the axis: x < 2 or x > 3.
Exactly. Less than zero means between the roots for an upward parabola; greater than zero means outside.
solve the related equation to find where the expression equals zero.
the expression can change sign only at a critical value, so one test point per interval is enough.
< and > exclude the critical values; ≤ and ≥ include them.
for a parabola opening upwards, < 0 holds between the roots and > 0 holds outside them.
multiply by −1 and reverse the inequality sign before using the graph rule.
use "or", as in x < 2 or x > 3, never "and".
Solve x² − 5x + 6 < 0: (x − 2)(x − 3) < 0, critical values 2 and 3, test points give +, −, +, so 2 < x < 3.
Solve x² − 5x + 6 ≥ 0: the same critical values, but now the positive intervals and the endpoints count, so x ≤ 2 or x ≥ 3.
Solve x² < 9: x² − 9 < 0, so (x − 3)(x + 3) < 0 and −3 < x < 3.
Solve −x² + 5x − 6 > 0: multiply by −1 and reverse the sign to get x² − 5x + 6 < 0, so 2 < x < 3.
Find the roots, then split the line; test each piece to check the sign; less than zero, the valley's low, between the roots is where you go.
A hiker walks across a field that dips into a valley between the 2 km and 3 km markers. Only between those markers is she below ground level, so that stretch is the solution of x² − 5x + 6 < 0.
"Less is in, greater is out" - for an upward parabola, < 0 lies between the roots and > 0 lies outside them.
C.T.W. - Critical values, Test points, Write the interval.
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What are the critical values of x² − 5x + 6 < 0?
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Solve x² − 5x + 6 < 0.
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Why are 2 and 3 excluded from the solution of x² − 5x + 6 < 0?
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For an upward parabola, where is y < 0?
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What must you do when multiplying an inequality by a negative number?
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An inequality involving a quadratic expression, such as x² − 5x + 6 < 0.
The solutions of the related equation, where the expression equals zero; here x = 2 and x = 3.
A value chosen inside an interval to find the sign of the expression there.
A table showing the sign of each factor, and of their product, on each interval.
An inequality using < or >, whose critical values are not part of the solution.
An inequality using ≤ or ≥, whose critical values are part of the solution.
A way of writing a range of values; 2 < x < 3 is written (2, 3), or ]2, 3[ in IB style.
Example 1 solves the quadratic inequality x² − 5x + 6 < 0. Factorising gives critical values 2 and 3, a sign check shows the expression is negative only between them, so the solution is 2 < x < 3.
A quadratic inequality asks for a range of x-values rather than single answers. For x² − 5x + 6 < 0, we first solve the related equation x² − 5x + 6 = 0 to find the critical values x = 2 and x = 3, which split the number line into three intervals. Testing one value in each interval, or using a sign table or the graph of y = x² − 5x + 6, shows that the expression is negative only between 2 and 3. Because the inequality is strict, the endpoints are left out, so the solution is 2 < x < 3. The lesson also shows how the answer changes for ≤, > and ≥, and how to handle a negative leading coefficient.
This is the first of two worked examples on quadratic inequalities. Example 2 solves 2x² − 8x + 6 ≥ 0, a non-strict inequality; here we solve a strict one and compare three ways of reaching the answer.
Example 1: Solve x² − 5x + 6 < 0.
Step 1: Find the critical values. Solve the related equation x² − 5x + 6 = 0. Two numbers that multiply to 6 and add to −5 are −2 and −3, so (x − 2)(x − 3) = 0 and the critical values are x = 2 and x = 3. At these points the expression equals 0.
Step 2: Split the number line. The critical values divide the number line into three intervals: x < 2, 2 < x < 3 and x > 3. In each interval the expression keeps the same sign, because it can only change sign where it passes through 0.
Step 3: Test one value in each interval.
Step 4: Write the solution. The expression is negative only between the critical values. The inequality is strict (< rather than ≤), and at x = 2 and x = 3 the expression equals 0, which is not less than 0. So the endpoints are excluded and the solution is 2 < x < 3. In interval notation this is (2, 3); IB papers also write it as ]2, 3[.
Two other ways to see the same answer.
How the answer changes with the sign.
A negative leading coefficient. −x² + 5x − 6 > 0 looks different, but multiplying both sides by −1 reverses the inequality sign: x² − 5x + 6 < 0. So it has the same solution, 2 < x < 3.
Common mistakes.