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$A=\left\{x\in N:x^2-5x-14=0\right\}$ হলে, $A =?$

Options :

  • (ক) $6,1$
  • (খ) $-2,7$
  • (গ) $2,7$
  • (ঘ) $7$

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এখানে,

$x^2 - 5x - 14 =0$

$\Rightarrow x^2 - 7x + 2x-14 = 0$

$\Rightarrow x(x-7) + 2(x-7)= 0$

$\Rightarrow (x-7)(x + 2) = 0$

$\therefore x = -2, 7$

কিন্তু $x \in N$ হওয়ায় $x$ এর মান $-2$ গ্রহণযোগ্য নয়। কারণ $N$ অর্থ স্বাভাবিক সংখ্যা ($N$ = Natural number) যা $1, 2, 3, \dots$ । এজন্য $x$ এর ঋণাত্মক মান নেওয়ার সুযোগ নেই।

$\therefore A = \{7\}$

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