Finding values without the zeros
Show on a big screenα and β are the zeros of x² − 4x + 1. Find α² + β². The zeros are not whole numbers. Finding them first takes much longer.
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(α + β)² = α² + 2αβ + β².(α + β)² = α² + 2αβ + β².- So
α² + β² = (α + β)² − 2αβ. - That is
S² − 2P. - For
S = 4andP = 1:16 − 2 = 14.
The rule
The answer is 14, found without α or β. Here S = α + β and P = αβ. Read S and P from the coefficients. Then write the asked value using S and P.
Why: (α + β)² = α² + 2αβ + β². Take 2αβ from both sides: α² + β² = (α + β)² − 2αβ.
`α² + β² = S² − 2P`.
For x² − 4x + 1: a = 1, b = −4 and c = 1. So S = 4 and P = 1. Then α² + β² = 16 − 2 = 14.
The reciprocal of α is 1/α. Reciprocals use the same method: 1/α + 1/β = β/(αβ) + α/(αβ) = (α + β)/(αβ) = S/P. This needs P not equal to 0, since 1/0 does not exist.
| Quadratic | S | P | S/P | Zeros | Check |
|---|---|---|---|---|---|
2x² − 12x + 10 | 6 | 5 | 6/5 | 1, 5 | 1 + 1/5 = 6/5 |
For x² − 4x + 1 we do not need α or β. The value S² − 2P is the whole answer.
Watch out!
Do not say α² + β² is just S². For zeros 2 and 3, S = 5 and S² = 25. But 4 + 9 = 13. Subtract 2P = 12: 25 − 12 = 13.
Try the lab to finish this lesson.