35 min readNo calculatorNeeds: factorising quadratics
Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — a quadratic inequality from start to finish
Solve x2−x−6>0.
1One side is already zero, so factorise: (x−3)(x+2)>0Always rearrange to zero before factorising.
2Critical values: x=3 and x=−2These are the values that make each bracket vanish.
3The coefficient of x2 is positive, so the parabola opens upwards.It dips below the axis between the roots and is above it outside.
4x<−2 or x>3We want where the curve is above the axis. Two regions, joined with 'or'.
5Check x=0: 0−0−6=−6, not greater than 0 — correctly excluded.One test value from each region confirms the direction.
Example 2 — rearranging before you factorise
Solve 2x2+5x≤3.
12x2+5x−3≤0Bring the 3 across. Factorising before rearranging gives nonsense.
2(2x−1)(x+3)≤0Two numbers multiplying to 2×−3=−6 and adding to 5: they are 6 and −1.
3Critical values: x=21 and x=−3Set each bracket to zero.
4Opens upwards, and we want ≤0, so we want the region between.Below or on the axis means between the roots.
5−3≤x≤21The inequality is not strict, so both endpoints are included.
Example 3 — a rational inequality, done safely
Solve x−1x+4≥2.
1x−1x+4−2≥0Subtract, never multiply. The sign of x−1 is unknown.
2x−1x+4−2(x−1)=x−16−x≥0Put it over one denominator: x+4−2x+2=6−x.
3Multiply by (x−1)2, which is always positive: (6−x)(x−1)≥0Squaring the denominator makes the multiplier safe, so the sign never flips.
4Critical values 6 and 1; the parabola (6−x)(x−1) opens downwards, so it is above the axis between them: 1≤x≤6The −x in the first bracket makes the leading coefficient negative.
5x=1 makes the original denominator zero, so exclude it: 1<x≤6Check x=7: 611≈1.83, which is less than 2 — correctly excluded.