৩য় অধ্যায়: বীজগাণিতিক রাশি
প্রশ্ন - ১:
a=√6+√5 এবং x=9+4√5।
ক. একটি ধনাত্মক সংখ্যা m এর বর্গ ঐ সংখ্যার পাঁচগুণের চেয়ে এক কম হলে, দেখাও যে, m+1m=5 ।
খ. প্রমাণ করো যে, x4−322x2+1=0 ।
গ. দেখাও যে, a12−1a6=1932√30
১ নং প্রশ্নের সমাধান
ক
দেওয়া আছে, একটি ধনাত্মক সংখ্যা m এর বর্গ ঐ সংখ্যার পাঁচগুণের চেয়ে এক কম। অর্থাৎ
m2=5m−1
⇒m2+1=5m
⇒m2m+1m=5mm [উভয় পক্ষে m দ্বারা ভাগ করে ]
∴ [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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