Let $P (x) = 24 x^{24} + \sum_{j = 1}^{23} (24 - j) (x^{24 - j} + x^{24 + j}) .$ Let $X_{1}, X_{2} ...... X_{r}$ be the distinct zeros of P(x) and let $x_{k}^{2} = a_{k} + i b_{k} f o r k = 1, 2,$ ……r where $a_{k}, b_{k}$ are real numbers, and $i^{2} = - 1$ . Let $\sum_{k = 1}^{r} | a_{k} | = m + n \sqrt{p}$ where m, n and p are integers and ‘p’ is prime then the value of m + n + p is equal to …...

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Detailed Solution

$P (x) = 24 x^{24} + 23 (x^{23} + x^{25}) + 22 (x^{22} + x^{26}) + ..... + 1 (x^{13} + x^{47}) = x + 2 x^{2} + ............ + 24 x^{24} + ......... + x^{47} = x {(1 + x + ..... + x^{23})}^{2} = \frac{x {(x^{24} - 1)}^{2}}{{(x - 1)}^{2}} P (x) = 0 \Rightarrow x^{24} = 1 x = 0, α, α^{2} ......... α^{23} w h e r e α = e^{i 2 π / 24} \sum_{K = 1}^{r} | a_{K} | = \sum_{K = 1}^{r} Re (| x_{k}^{2} |) = 7 + 4 \sqrt{3}$