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Algebra · Inequalities

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Quadratic inequalities

Why x² < 16 has TWO boundaries, how the parabola sketch decides "between" versus "outside", and why two-region answers must be written with "or" — never chained into one impossible inequality.

Algebra · Inequalities

Quadratic inequalities

Why x² < 16 has TWO boundaries, how the parabola sketch decides "between" versus "outside", and why two-region answers must be written with "or" — never chained into one impossible inequality.

Why it works

x2≤16x^2 \le 16 tempts a one-line answer: "x≤4x \le 4". But test x=−10x = -10: 100≤16100 \le 16 is false — yet −10≤4-10 \le 4. Squaring wipes out signs, so BIG negative numbers have big squares too. The truth has two boundaries:

−4≤x≤4.-4 \le x \le 4.

The routine: critical values first, then a sketch decides the region.
  1. Move everything to one side: quadratic≶0\text{quadratic} \lessgtr 0.
  2. Factorise to find the critical values — where the quadratic IS
zero.
  1. Sketch the parabola (a happy U when the x2x^2 coefficient is
positive): it dips below zero BETWEEN its roots and sits above zero OUTSIDE them.
  1. Read off the region the inequality asks for.
For x2−3x−10>0x^2 - 3x - 10 > 0: factorise (x−5)(x+2)(x - 5)(x + 2), critical values −2-2 and 55. "Greater than 0" wants the parabola ABOVE the axis — outside the roots:

x<−2orx>5.x < -2 \quad\text{or}\quad x > 5.

For x2+x−12<0x^2 + x - 12 < 0: (x+4)(x−3)(x + 4)(x - 3), and "below zero" is between: −4<x<3-4 < x < 3.

Two regions need the word "or". Writing −2>x>5-2 > x > 5 chains the regions into one statement claiming −2>5-2 > 5 — impossible, and it loses the mark. Between-answers chain happily (−4<x<3-4 < x < 3); outside-answers never do.

Don't divide by xx. x2>3xx^2 > 3x invites "÷x: x>3x > 3" — but xx might be negative (flipping the symbol) or zero (illegal). Bring everything to one side instead: x(x−3)>0x(x - 3) > 0, critical values 0 and 3, outside region: x<0x < 0 or x>3x > 3. Test x=−1x = -1: 1>−31 > -3. ✓

Whole-number twists. "n is an integer with n2<30n^2 < 30": the boundary is 30≈5.48\sqrt{30} \approx 5.48, so −5.48<n<5.48-5.48 < n < 5.48 — greatest integer 5, least −5-5. The square root is never a licence to forget the negative side.