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,
\(\ \ \ \ x=9+4\sqrt{5}\)
\(\therefore \ \dfrac{1}{\sqrt{x}}=\dfrac{1}{(\sqrt{5}+2)}\)
\(\therefore \ \ \ \ \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=(\sqrt{5}+2)+(\sqrt{5}-2)\)
\(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=\sqrt{5}+2+\sqrt{5}-2\)
\(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=2\sqrt{5}\)
\(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2=(2\sqrt{5})^2\) [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]
\(\ \ \ \ x=9+4\sqrt{5}\)
\(=5+4\sqrt{5}+4\)
\(=(\sqrt{5})^2+2.2\sqrt{5}+(2)^2\)
\(=(\sqrt{5}+2)^2\)
\(\therefore \ \dfrac{1}{\sqrt{x}}=\dfrac{1}{(\sqrt{5}+2)}\)
\(=\dfrac{(\sqrt{5}-2)}{(\sqrt{5}+2)(\sqrt{5}-2})\)
\(=\dfrac{(\sqrt{5}-2)}{(\sqrt{5})^2-(2)^2}\)
\(=\dfrac{(\sqrt{5}-2)}{(5-4)}\)
\(\therefore \ \ \ \ \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=(\sqrt{5}+2)+(\sqrt{5}-2)\)
\(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=\sqrt{5}+2+\sqrt{5}-2\)
\(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=2\sqrt{5}\)
\(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2=(2\sqrt{5})^2\) [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,
\(\ \ \ \ x=9+4\sqrt{5}\)
\(\therefore \ \dfrac{1}{\sqrt{x}}=\dfrac{1}{(\sqrt{5}+2)}\)
\(\therefore \ \ \ \ \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=(\sqrt{5}+2)+(\sqrt{5}-2)\)
\(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=\sqrt{5}+2+\sqrt{5}-2\)
\(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=2\sqrt{5}\)
\(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2=(2\sqrt{5})^2\) [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]
\(\ \ \ \ x=9+4\sqrt{5}\)
\(=5+4\sqrt{5}+4\)
\(=(\sqrt{5})^2+2.2\sqrt{5}+(2)^2\)
\(=(\sqrt{5}+2)^2\)
\(\therefore \ \dfrac{1}{\sqrt{x}}=\dfrac{1}{(\sqrt{5}+2)}\)
\(=\dfrac{(\sqrt{5}-2)}{(\sqrt{5}+2)(\sqrt{5}-2})\)
\(=\dfrac{(\sqrt{5}-2)}{(\sqrt{5})^2-(2)^2}\)
\(=\dfrac{(\sqrt{5}-2)}{(5-4)}\)
\(\therefore \ \ \ \ \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=(\sqrt{5}+2)+(\sqrt{5}-2)\)
\(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=\sqrt{5}+2+\sqrt{5}-2\)
\(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)=2\sqrt{5}\)
\(\Rightarrow \left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2=(2\sqrt{5})^2\) [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]
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