Split and solve
Show on a big screenSplitting the middle term, bx, can solve a quadratic equation.
2x² − 7x + 3 = 0: a = 2, b = −7, c = 3. Need product a·c = 6, sum b = −7.2x² − 7x + 3 = 0: a = 2, b = −7, c = 3. Need product a·c = 6, sum b = −7.−1and−6: product 6, sum −7.- Split:
2x² − x − 6x + 3 = 0. - Group and bracket:
(2x − 1)(x − 3) = 0.
The rule
Four moves solve it when move 2 finds its two numbers.
- Tidy to standard form.
- Split the middle term. Find two numbers with product
a·cand sumb. Herea·cmeans a times c. - Bracket the four terms in two groups.
- Set each bracket to 0.
A product is 0 only when a factor is 0. So set each bracket to 0 in turn.
Group the first two terms: 2x² − x = x(2x − 1). Group the last two: −6x + 3 = −3(2x − 1). Both groups hold the bracket (2x − 1). Take it out: (2x − 1)(x − 3).
Why product a·c? Suppose the left side is two brackets multiplied. Then the two split terms always multiply to a·c. Here (−1) × (−6) = 6, and 2 × 3 = 6. So search for that product.
2x − 1 = 0 gives x = 1/2. x − 3 = 0 gives x = 3.
Check x = 3: 2 × 9 − 21 + 3 = 18 − 21 + 3 = 0. Check x = 1/2: 2x² is 2 × 1/4 = 1/2, and 7x is 7/2. So 1/2 − 7/2 + 3 = −3 + 3 = 0.
Some quadratics have no whole-number pair. Then use the formula.
| Move | Result |
|---|---|
| Tidy | 2x² − 7x + 3 = 0 |
| Split | 2x² − x − 6x + 3 = 0 |
| Bracket | (2x − 1)(x − 3) = 0 |
| Set to 0 | x = 1/2 or x = 3 |
Watch out!
Dividing x² = 3x by x gives x = 3. It loses x = 0, which also balances. Tidy to x(x − 3) = 0. Both roots stay.
Now try it. Tap the four moves in order.
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