An independent check is a second route to the same result, chosen so a slip in the first route is unlikely to repeat. It is the last habit in advanced non-calculator reasoning, and it protects the marks you have already earned.
Which checks are cheap and different?
- Substitute the answer back into the original equation, not your rearranged version.
- Sum and product of roots: for ax² + bx + c = 0, the roots add to −b/a and multiply to c/a.
- Estimate with friendly decimals, such as √3 ≈ 1.73, to see if the size is sensible.
- Test a simple value on both sides of an identity, such as x = 1 or x = 2.
Pick one that does not reuse your earlier steps.
Worked example
Solve 3x² − 5x − 2 = 0, then check.
Method 1, factorise: 3x² − 5x − 2 = (3x + 1)(x − 2). Setting each bracket to zero gives x = −1/3 and x = 2.
Check A, sum of roots: the roots should add to −b/a = 5/3. Here −1/3 + 2 = 5/3. ✓
Check B, product of roots: the roots should multiply to c/a = −2/3. Here (−1/3) × 2 = −2/3. ✓
Check C, substitute x = 2: 3(4) − 10 − 2 = 12 − 12 = 0. ✓
Check D, the formula: x = (5 ± √(25 + 24))/6 = (5 ± 7)/6, which gives 2 and −1/3. ✓
Checks A and B take about ten seconds and do not reuse the factorising, so they are independent of Method 1.
The mistake to watch for
A common slip is a flipped sign that the same method cannot see.
Mistaken answer: x = −2 and x = 1/3
The student wrote the brackets correctly but set them equal to zero with the wrong signs.
Redoing the same steps would very likely give the same mistake. The sum of roots exposes it at once: −2 + 1/3 = −5/3, but the equation needs +5/3. The mismatch tells the student to look again.
Check yourself
Try these without a calculator, then open each answer.
1. A student says the roots of x² − 7x + 12 = 0 are 3 and 4. Check using the sum and product.
Show answer
Sum should be 7 and product 12. Here 3 + 4 = 7 and 3 × 4 = 12. Both agree, so the roots are correct.
2. Solve √(x + 6) = x, then check each answer in the original equation.
Show answer
Squaring: x + 6 = x², so x² − x − 6 = 0, which gives (x − 3)(x + 2) = 0. Test x = 3: √9 = 3 ✓. Test x = −2: √4 = 2, but the right side is −2, so ✗. The only solution is x = 3.
3. Expand (√5 + 1)² and check with √5 ≈ 2.236.
Show answer
(√5 + 1)² = 5 + 2√5 + 1 = 6 + 2√5. Check: 6 + 4.472 = 10.472, and (3.236)² = 10.472. They agree.
Where this leads next
Put all five lessons together in the mixed practice set, and record any slips in the mistake log. The quadratic structure explorer and the non-calculator working trainer help you test a result afterwards.
Some students know how to check but skip it under pressure. Our teachers help you build a checking routine that fits inside the time you have in online one-to-one Additional Mathematics tuition.