Given:
\[ 15(2A^2-B^2)=7AB \]
\[ 30A^2-15B^2=7AB \]
Divide by \(B^2\):
\[ 30\left(\frac{A}{B}\right)^2-15=7\left(\frac{A}{B}\right) \]
Let \(\frac{A}{B}=x\)
\[ 30x^2-7x-15=0 \]
\[ (5x+3)(6x-5)=0 \]
Since \(A,B\) are positive, \(x=\frac{5}{6}\)
\[ \frac{A}{B}=\frac{5}{6} \]
Answer: 5 : 6
দেওয়া আছে:
\[ 15(2A^2-B^2)=7AB \]
\[ 30A^2-15B^2=7AB \]
উভয় পাশে \(B^2\) দ্বারা ভাগ করি:
\[ 30\left(\frac{A}{B}\right)^2-15=7\left(\frac{A}{B}\right) \]
ধরি \(\frac{A}{B}=x\)
\[ 30x^2-7x-15=0 \]
\[ (5x+3)(6x-5)=0 \]
যেহেতু \(A,B\) ধনাত্মক, তাই \(x=\frac{5}{6}\)
অতএব,
\[ A:B=5:6 \]
উত্তর: 5 : 6
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