The pq formula: solving quadratic equations

We want to find the values of x that make the left-hand side zero. Here p = −6 and q = 8.

1. We put in p and q. The signs stay with the numbers.

   x = −(−6)/2 ± √((−$\frac{6}{2}$)² − 8)

2. −(−6)/2 = 3 and (−3)² = 9, because minus times minus is plus.

   x = 3 ± √(9 − 8)

3. 9 − 8 = 1 and √1 = 1.

   x = 3 ± 1

4. With plus we get 4, with minus 2.

   x = 4 or x = 2

x = 4: 16 − 24 + 8 = 0. x = 2: 4 − 12 + 8 = 0. Both give zero, so both are correct.

We want x² with no number in front. We divide every term on both sides by 2.

1. 2x² ÷ 2 = x², 4x ÷ 2 = 2x, −30 ÷ 2 = −15 and 0 ÷ 2 = 0.

   x² + 2x − 15 = 0

2. Now p = 2 and q = −15. So −p/2 = −1 and p/2 = 1.

   x = −1 ± √(1² − (−15))

3. Minus minus 15 is plus 15, so 1 + 15 = 16.

   x = −1 ± √16

4. The square root of 16 is 4.

   x = −1 ± 4

5. −1 + 4 = 3 and −1 − 4 = −5.

   x = 3 or x = −5

x = 3: 18 + 12 − 30 = 0. x = −5: 50 − 20 − 30 = 0. Both are correct in the original equation.

We want to find x. Here p = −4 and q = −1.

1. −p/2 = 2 and (p/2)² = 4. Minus q becomes plus 1.

   x = 2 ± √(4 + 1)

2. √5 is not a whole number, so we leave the root as it is and give an exact answer.

   x = 2 ± √5

3. As an approximation, √5 ≈ 2.236.

   x ≈ 4.24 or x ≈ −0.24

The approximations: 4.236² − 4 · 4.236 − 1 ≈ 0.00 and (−0.236)² − 4 · (−0.236) − 1 ≈ 0.00. Both are correct.