Further Inequalities [H]

Further Inequalities

At Higher tier, inequalities extend to two dimensions and to quadratic expressions. Representing inequalities graphically and solving quadratic inequalities both require careful attention to the sign of a quadratic expression across different regions.

Linear Inequalities in Two Variables

An inequality like \( 2x + y leq 6 \) is satisfied by all points on one side of the line \( 2x+y=6 \). To identify which side:

  1. Draw the boundary line \( 2x+y=6 \) (solid for \( leq \) or \( geq \); dashed for \( < \) or \( > \)).
  2. Test a point not on the line (the origin \((0,0)\) is usually easiest): \( 2(0)+0 = 0 leq 6 \) ✓ — so the origin side is satisfied.
  3. Shade the region that is not satisfied (some conventions) or shade the satisfied region — follow the instruction in the question.

The intersection of several half-planes gives the feasible region satisfying all inequalities simultaneously.

Quadratic Inequalities

To solve \( ax^2 + bx + c > 0 \) (or \( < 0 \)): find the roots of \( ax^2+bx+c = 0 \), then determine the sign of the quadratic in each interval between the roots.

  • For \( a > 0 \) (U-shape): the quadratic is negative between the roots and positive outside.
  • For \( a < 0 \) (∩-shape): the quadratic is positive between the roots and negative outside.
\[ x^2 - 5x + 6 > 0 \implies (x-2)(x-3)>0 \implies x < 2 \text{ or } x > 3 \] \[ x^2 - 5x + 6 < 0 \implies 2 < x < 3 \]

Worked Examples

Represent the region satisfying \( x + 2y leq 8 \), \( x > 1 \) and \( y geq 0 \) on a coordinate grid.

Draw three boundary lines: \(x+2y=8\) (solid), \(x=1\) (dashed), \(y=0\) (solid x-axis). Test origin: \(0+0=0leq 8\) ✓, \(0>1\) ✗ — so \(x>1\) excludes the origin side. The feasible region is bounded on the left by \(x=1\), above by \(x+2y=8\) and below by \(y=0\).

Solve \( x^2 - x - 6 > 0 \).

Roots: \( (x-3)(x+2)=0 Rightarrow x=3 \) or \( x=-2 \). Since \( a=1>0 \), the parabola is U-shaped — positive outside the roots.

\[ x < -2 \text{ or } x > 3 \]
Solve \( x^2 - x - 6 \leq 0 \).

Same roots; now want where the parabola is at or below zero: between the roots.

\[ -2 \leq x \leq 3 \]
Solve \( -x^2 + 4x > 0 \).

Roots: \( -x^2+4x = -x(x-4) = 0 Rightarrow x=0 \) or \( x=4 \). Since \( a=-1<0 \), ∩-shape — positive between the roots.

\[ 0 < x < 4 \]

 Key Takeaways

  • Two-variable inequalities: draw the boundary line (solid for \(leq/geq\), dashed for \()\), test a point, shade the required region.
  • Quadratic inequality: find the roots, then use the parabola shape (\(a>0\): negative between roots; \(a<0\): positive between roots).
  • For \(ax^2+bx+c>0\) with \(a>0\): solution is \(x < r_1\) or \(x > r_2\) (outside the roots).
  • For \(ax^2+bx+c<0\) with \(a>0\): solution is \(r_1 < x < r_2\) (between the roots).
  • Always sketch the parabola to identify the correct region — do not rely on memory alone for the sign pattern.