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If \(x^{2}-\frac{1}{x^{2}}=24\) and \(x-\frac{1}{x}=4\), what is the value of \(x+\frac{1}{x}\)?यदि \(x^{2}-\frac{1}{x^{2}}=24\) और \(x-\frac{1}{x}=4\), तो \(x+\frac{1}{x}\) का मान क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (6)(6)
Since \(x^{2}-\frac{1}{x^{2}}=\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\), \(24=4\left(x+\frac{1}{x}\right)\). In exams, use the difference of squares identity.\(x^{2}-\frac{1}{x^{2}}=\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\), इसलिए \(24=4\left(x+\frac{1}{x}\right)\)। परीक्षा में वर्गों के अंतर की पहचान लगाएं।
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