Verified by Ajay Kumar
Find the value of \(\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}\).
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Answer: B
Let \(x=\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}\). Since the expression repeats indefinitely, \(x=\sqrt{12+x}\). Squaring gives \(x^2=12+x\), so \(x^2-x-12=0\). Factoring gives \((x-4)(x+3)=0\). Since a square root is non-negative, \(x=4\). Trick: Let the repeating radical equal \(x\) and solve the resulting quadratic equation.
\(\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}\) का मान ज्ञात कीजिए।
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उत्तर: B
मान लें \(x=\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}\)। चूँकि यह व्यंजक अनंत बार दोहराया गया है, इसलिए \(x=\sqrt{12+x}\)। वर्ग करने पर \(x^2=12+x\), अर्थात \(x^2-x-12=0\)। गुणनखंड करने पर \((x-4)(x+3)=0\) मिलता है। वर्गमूल ऋणात्मक नहीं होता, इसलिए \(x=4\)। ट्रिक: दोहराए जाने वाले वर्गमूल को \(x\) मानकर प्राप्त द्विघात समीकरण हल करें।
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