L5 · Quadratics and inequalities clinic

Learning goal

“I can move between the unit's two families without prompting: isolate-and-root for x² = a and x² + c = 0 (counting solutions correctly), and solve-and-represent for linear inequalities (reversing only on a negative multiplier/divisor) - verifying everything by substitution or test values.”

Plus-or-minus (±) — both signed roots travel together whenever a > 0 (from L1).

No real solution — the complete, provable answer when a square is set equal to a negative (from L1-L2).

Boundary value — where equality would hold; included exactly when the symbol carries its bar (from L3).

Reversal — the symbol flips when, and only when, both sides are multiplied or divided by a negative (from L4).

Worked example

Solve x² - 60 = 0: isolate: x² = 60, so x = ±√60 (two solutions, exact form). Solve x² + 13 = 5: isolate: x² = -8, so no real solution. Solve x² + 7 = 7: isolate: x² = 0, so x = 0, ONE solution - the ± pair collapses when the target is 0. Say it plainly: every quadratic in this unit runs the same two beats - isolate, then read the target's sign. Positive: a ± pair. Zero: the single value 0. Negative: no real solution. The form of the question changes; the beats don't.

Check yourself — explain it

x² + 7 = 7 has exactly one solution because...

Paired problem — example still visible

Solve x² - 29 = 0 -> x² = 29, so x = ±√29 (two solutions)

Faded completion — last step is yours

"x² + 3 = 12: isolate to get x² = 9, then x = ± ___" -> 3

Ready to practise?

10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.

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