how-is-the-solution-of-this-qution-x-x-1-x-2-x-3-1-when-determinant-x-50-determinant-

Question Number 175715 by zaheen last updated on 05/Sep/22

h o w i s t h e s o l u t i o n o f t h i s q u t i o n

\sqrt{(x) (x + 1) (x + 2) (x + 3) + 1}

w h e n \begin{matrix} x = 50 \end{matrix} \begin{array}{cc}  \end{array}

Commented by Frix last updated on 05/Sep/22

(x + n) (x + n + 1) (x + n + 2) (x + n + 3) + 1 =

= {(x^{2} + (2 n + 3) x + n^{2} + 3 n + 1)}^{2}

Answered by ajfour last updated on 05/Sep/22

l e t x = t - \frac{3}{2} \Rightarrow x + 3 = t + \frac{3}{2}

f (t) = \sqrt{(t^{2} - \frac{9}{4}) (t^{2} - \frac{1}{4}) + 1}

l e t t^{2} - \frac{5}{4} = s

f (s) = \sqrt{s^{2} - 1 + 1} = s

= t^{2} - \frac{5}{4} = {(x + \frac{3}{2})}^{2} - \frac{5}{4}

f (50) = \frac{103^{2} - 5}{4} = 2651

Commented by Tawa11 last updated on 05/Sep/22

Great sir

Answered by behi834171 last updated on 05/Sep/22

x (x + 3) = x^{2} + 3 x

(x + 1) (x + 2) = x^{2} + 3 x + 2

x^{2} + 3 x = t

\Rightarrow f (t) = \sqrt{(t) (t + 2) + 1} = \sqrt{t^{2} + 2 t + 1} = t + 1

\Rightarrow f (x) = x^{2} + 3 x + 1

\Rightarrow f (x) = 50^{2} + 3 \times 50 + 1 = 2651 .