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Step-by-Step Solution
Step 1: Identify the quadratic equation and its roots
We are given the quadratic equation
$ x^2 - x - 1 = 0 $
with roots $ \alpha $ and $ \beta $. By Viète’s relations:
$ \alpha + \beta = 1 $ and $ \alpha \beta = -1 $.
We also define
$ p_k = \alpha^k + \beta^k.
$
Step 2: Use the relation $ \alpha^2 = \alpha + 1 $
From the equation $ \alpha^2 - \alpha - 1 = 0 $, we can write:
$ \alpha^2 = \alpha + 1.
$
Step 3: Express higher powers of $ \alpha $
(a) For $ \alpha^3 $:
$ \alpha^3 = \alpha \cdot \alpha^2 = \alpha (\alpha + 1) = \alpha^2 + \alpha = (\alpha + 1) + \alpha = 2\alpha + 1.
$
(b) For $ \alpha^4 $:
$ \alpha^4 = \alpha \cdot \alpha^3 = \alpha (2\alpha + 1) = 2\alpha^2 + \alpha = 2(\alpha + 1) + \alpha = 3\alpha + 2.
$
(c) For $ \alpha^5 $:
$ \alpha^5 = \alpha \cdot \alpha^4 = \alpha (3\alpha + 2) = 3\alpha^2 + 2\alpha.
$
Using $ \alpha^2 = \alpha + 1 $ again:
$ \alpha^5 = 3(\alpha + 1) + 2\alpha = 3\alpha + 3 + 2\alpha = 5\alpha + 3.
$
Step 4: Find $ p_5 = \alpha^5 + \beta^5 $
Since $ p_k = \alpha^k + \beta^k $, we have
$ p_5 = \alpha^5 + \beta^5.
$
By symmetry and using the same process for $ \beta $, we get
$ \beta^5 = 5\beta + 3.
$
Hence,
$ p_5 = (5\alpha + 3) + (5\beta + 3) = 5(\alpha + \beta) + 6.
$
Because $ \alpha + \beta = 1 $, we get
$ p_5 = 5 \cdot 1 + 6 = 11.
$
Step 5: Compute $ p_2 \cdot p_3 $
We know:
$ p_2 = \alpha^2 + \beta^2, \quad p_3 = \alpha^3 + \beta^3.
$
Using the fact that
$ \alpha^2 = \alpha + 1, \; \beta^2 = \beta + 1,
$ we get:
$ p_2 = (\alpha + 1) + (\beta + 1) = (\alpha + \beta) + 2 = 1 + 2 = 3.
$
Next,
$ \alpha^3 = 2\alpha + 1, \; \beta^3 = 2\beta + 1,
$ so
$ p_3 = (2\alpha + 1) + (2\beta + 1) = 2(\alpha + \beta) + 2 = 2 \cdot 1 + 2 = 4.
$
Therefore,
$ p_2 \cdot p_3 = 3 \times 4 = 12.
$
Step 6: Check the statement $ p_5 = p_2 \cdot p_3 $
From the above results:
$ p_5 = 11 \quad \text{and} \quad p_2 \cdot p_3 = 12.
$
Thus,
$ p_5 \ne p_2 \cdot p_3.
$
Hence, the statement “$ p_5 = p_2 \cdot p_3 $” is not true.
Final Conclusion
The incorrect statement among the given options is indeed
$ p_5 = p_2 \cdot p_3,
$ because we have just shown it does not hold.
