L2 · Solving x² + c = 0 (c an integer)

Learning goal

“I can solve equations of the form x² + c = 0 (c an integer) by first isolating x², then applying the x² = a method - including recognizing when the isolated equation has no real solution - and verify by substitution.”

Isolate — to get the squared term alone on one side: in x² - 49 = 0, isolating gives x² = 49.

Constant term — the term with no variable: in x² + 16 = 0, the constant term is 16.

Rearranged form — the same equation after a move: x² + c = 0 rearranges to x² = -c; nothing about the solutions changes.

Check by substitution — replacing x with a claimed solution in the ORIGINAL equation and confirming both sides match (the habit that catches sign errors).

Worked example

Solve x² - 49 = 0. Step 1: Isolate the squared term: add 49 to both sides - x² = 49. Step 2: This is now exactly an L1 equation: x = ±√49 = ±7. Step 3: Verify in the ORIGINAL (not the rearranged line): 7² - 49 = 49 - 49 = 0. (-7)² - 49 = 49 - 49 = 0. Say it plainly: x² + c = 0 is x² = a wearing a coat. The first move is always the same - move the constant across and change its sign - and after that move, L1 finishes the job.

Check yourself — explain it

The first step in solving x² - 49 = 0 is to add 49 to both sides because...

Paired problem — example still visible

Solve x² - 25 = 0 and verify both solutions -> x² = 25, so x = ±5; check: 25 - 25 = 0 for both 5 and -5

Faded completion — last step is yours

"x² + 9 = 34: subtract 9 from both sides to get x² = 25, then x = ± ___" -> 5

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

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