Math Resolution For SSC Chapter-3

৩য় অধ্যায়: বীজগাণিতিক রাশি

প্রশ্ন - ১:

\(a=\sqrt{6}+\sqrt{5}\) এবং \(x=9+4\sqrt{5}\)।

ক. একটি ধনাত্মক সংখ্যা \(m\) এর বর্গ ঐ সংখ্যার পাঁচগুণের চেয়ে এক কম হলে, দেখাও যে, \(m+\dfrac{1}{m}=5\) ।

খ. প্রমাণ করো যে, \(x^4-322x^2+1=0\) ।

গ. দেখাও যে, \(\dfrac{a^{12}-1}{a^6}=1932\sqrt{30} \)

১ নং প্রশ্নের সমাধান

ক

দেওয়া আছে, একটি ধনাত্মক সংখ্যা \(m\) এর বর্গ ঐ সংখ্যার পাঁচগুণের চেয়ে এক কম। অর্থাৎ

\(\ \ \ \ m^2=5m-1\)
\(\Rightarrow m^2+1=5m\)
\(\Rightarrow \dfrac{m^2}{m}+\dfrac{1}{m}=\dfrac{5m}{m}\)   [উভয় পক্ষে \(m\) দ্বারা ভাগ করে ]
\(\therefore \ m+\dfrac{1}{m}=5\)   [দেখানো হলো ]

খ

দেওয়া আছে,
  \(\ \ \ \ x=9+4\sqrt{5}\)

  \(=5+4\sqrt{5}+4\)
  \(=(\sqrt{5})^2+2.2\sqrt{5}+(2)^2\)
  \(=(\sqrt{5}+2)^2\)

\(\therefore\ \sqrt{x}=(\sqrt{5}+2)\)
\(\therefore \ \dfrac{1}{\sqrt{x}}=\dfrac{1}{(\sqrt{5}+2)}\)

  \(=\dfrac{(\sqrt{5}-2)}{(\sqrt{5}+2)(\sqrt{5}-2})\)
  \(=\dfrac{(\sqrt{5}-2)}{(\sqrt{5})^2-(2)^2}\)
  \(=\dfrac{(\sqrt{5}-2)}{(5-4)}\)

\(\therefore \ \dfrac{1}{\sqrt{x}}=(\sqrt{5}-2)\)
\(\therefore \ \ \ \ \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=(\sqrt{5}+2)+(\sqrt{5}-2)\)
  \(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=\sqrt{5}+2+\sqrt{5}-2\)
  \(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=2\sqrt{5}\)
  \(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2=(2\sqrt{5})^2\)   [বর্গ করে]
  \(\Rightarrow (\sqrt{x})^2+2.\sqrt{x}.\dfrac{1}{\sqrt{x}}+\left(\dfrac{1}{\sqrt{x}}\right)^2=4\times5\)
  \(\Rightarrow (x+\dfrac{1}{x}=18 \cdots\ \cdots\ \cdots\ \) ①
  \(\Rightarrow \left(x+\dfrac{1}{x}\right)^2=(18)^2\)   [বর্গ করে]
  \(\Rightarrow (x)^2+2.x.\dfrac{1}{x}+\left(\dfrac{1}{x}\right)^2=324\)
  \(\Rightarrow x^2+\dfrac{1}{x^2}=324-2\)
  \(\Rightarrow \dfrac{(x)^4+1}{x^2}=322\)
  \(\Rightarrow x^4+1=322 x^2\)
  \(\therefore \ x^4-322 x^2+1=0\)   [প্রমাণিত]

গ

দেওয়া আছে,
  \(\ \ \ \ x=9+4\sqrt{5}\)

  \(=5+4\sqrt{5}+4\)
  \(=(\sqrt{5})^2+2.2\sqrt{5}+(2)^2\)
  \(=(\sqrt{5}+2)^2\)

\(\therefore\ \sqrt{x}=(\sqrt{5}+2)\)
\(\therefore \ \dfrac{1}{\sqrt{x}}=\dfrac{1}{(\sqrt{5}+2)}\)

  \(=\dfrac{(\sqrt{5}-2)}{(\sqrt{5}+2)(\sqrt{5}-2})\)
  \(=\dfrac{(\sqrt{5}-2)}{(\sqrt{5})^2-(2)^2}\)
  \(=\dfrac{(\sqrt{5}-2)}{(5-4)}\)

\(\therefore \ \dfrac{1}{\sqrt{x}}=(\sqrt{5}-2)\)
\(\therefore \ \ \ \ \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=(\sqrt{5}+2)+(\sqrt{5}-2)\)
  \(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=\sqrt{5}+2+\sqrt{5}-2\)
  \(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=2\sqrt{5}\)
  \(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2=(2\sqrt{5})^2\)   [বর্গ করে]
  \(\Rightarrow (\sqrt{x})^2+2.\sqrt{x}.\dfrac{1}{\sqrt{x}}+\left(\dfrac{1}{\sqrt{x}}\right)^2=4\times5\)
  \(\Rightarrow (x+\dfrac{1}{x}=18 \cdots\ \cdots\ \cdots\ \) ①
  \(\Rightarrow \left(x+\dfrac{1}{x}\right)^2=(18)^2\)   [বর্গ করে]
  \(\Rightarrow (x)^2+2.x.\dfrac{1}{x}+\left(\dfrac{1}{x}\right)^2=324\)
  \(\Rightarrow x^2+\dfrac{1}{x^2}=324-2\)
  \(\Rightarrow \dfrac{(x)^4+1}{x^2}=322\)
  \(\Rightarrow x^4+1=322 x^2\)
  \(\therefore \ x^4-322 x^2+1=0\)   [প্রমাণিত]