Zeroes and coefficients

Take 2x² – 8x + 6. Split the middle term to factorise it: 2x² – 6x – 2x + 6 = 2(x – 1)(x – 3). Its zeroes are 1 and 3.

Now look at the numbers a = 2, b = –8, c = 6:

• sum of zeroes = 1 + 3 = 4, and –b/a = –(–8)/2 = 4
• product of zeroes = 1 × 3 = 3, and c/a = 6/2 = 3

That is no accident. For every quadratic ax² + bx + c with zeroes α and β (Greek letters, 'alpha' and 'beta'):

Why it always works (3 lines): if α and β are the zeroes, then (x – α) and (x – β) are factors, so

ax² + bx + c = a(x – α)(x – β) = ax² – a(α + β)x + aαβ. Match the parts: b = –a(α + β) and c = aαβ. So α + β = –b/a and αβ = c/a.

Two more checks:

• x² + 7x + 10 = (x + 2)(x + 5): zeroes –2 and –5, sum –7 = –7/1, product 10 = 10/1.
• x² – 3 = (x – √3)(x + √3): zeroes √3 and –√3, sum 0 (there is no x term, so b = 0), product –3.

A habit that saves marks: before anything else, write down a, b and c, with their signs. For 6x² – 3 – 7x, first put it in order: 6x² – 7x – 3, so a = 6, b = –7, c = –3. Write the formula's minus sign first, then put b in: –(–7)/6 = 7/6.