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If \(r=\sqrt{21}+\sqrt{14}\), what is the value of \(r^2-14\sqrt{6}\)?यदि \(r=\sqrt{21}+\sqrt{14}\) है, तो \(r^2-14\sqrt{6}\) का मान क्या होगा?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: 3535
Using the square of a binomial, \(r^2=(\sqrt{21}+\sqrt{14})^2=21+14+2\sqrt{294}\). Since \(\sqrt{294}=\sqrt{49\times6}=7\sqrt{6}\), we get \(r^2=35+14\sqrt{6}\). Therefore, \(r^2-14\sqrt{6}=35\), so option C is correct. Exam tip: In such problems, simplify the cross-term \(2\sqrt{ab}\) carefully before subtracting.द्विपद के वर्ग का सूत्र लगाएँ: \(r^2=(\sqrt{21}+\sqrt{14})^2=21+14+2\sqrt{294}\)। क्योंकि \(\sqrt{294}=\sqrt{49\times6}=7\sqrt{6}\), इसलिए \(r^2=35+14\sqrt{6}\)। अतः \(r^2-14\sqrt{6}=35\), इसलिए विकल्प C सही है। परीक्षा-युक्ति: ऐसे प्रश्नों में पहले \(2\sqrt{ab}\) वाले पद को सरल करें।
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