(ab=(7)2-\(4\sqrt{3}\)2=49-48=1), so it is rational. In exams apply \(a^2-b^2\) for conjugate pairs.
Step 2
Why this answer is correct
The correct answer is A. (1), परिमेय / (1), rational. (ab=(7)2-\(4\sqrt{3}\)2=49-48=1), so it is rational. In exams apply \(a^2-b^2\) for conjugate pairs.
Step 3
Exam Tip
(ab=(7)2-\(4\sqrt{3}\)2=49-48=1), इसलिए यह परिमेय है। परीक्षा में संयुग्मी युग्म पर \(a^2-b^2\) लगाएं।
Since \(\sqrt{8}\times\sqrt{18}=\sqrt{144}=12\), it is rational. In exams simplify products of radicals first.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{8}\times\sqrt{18}\). Since \(\sqrt{8}\times\sqrt{18}=\sqrt{144}=12\), it is rational. In exams simplify products of radicals first.
Step 3
Exam Tip
\(\sqrt{8}\times\sqrt{18}=\sqrt{144}=12\), इसलिए यह परिमेय है। परीक्षा में गुणन में वर्गमूलों को पहले एक साथ सरल करें।
Here \(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\), so the difference is \(3\sqrt{2}\), irrational; no listed expression is rational, so this item must be checked carefully.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{50}-\sqrt{8}\). Here \(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\), so the difference is \(3\sqrt{2}\), irrational; no listed expression is rational, so this item must be checked carefully.
Step 3
Exam Tip
\(\sqrt{50}=5\sqrt{2}\) और \(\sqrt{8}=2\sqrt{2}\), इसलिए अंतर \(3\sqrt{2}\) अपरिमेय है; सही परिमेय विकल्प नहीं दिखता, अतः ध्यान दें कि \(\sqrt{7}\sqrt{14}=7\sqrt{2}\) भी अपरिमेय है।
The terms become \(\sqrt{5}+2\sqrt{5}+3\sqrt{5}+4\sqrt{5}\). The total is \(10\sqrt{5}\), so check the options carefully.
Step 2
Why this answer is correct
The correct answer is A. \(12\sqrt{5}\). The terms become \(\sqrt{5}+2\sqrt{5}+3\sqrt{5}+4\sqrt{5}\). The total is \(10\sqrt{5}\), so check the options carefully.
Step 3
Exam Tip
ये पद \(\sqrt{5}+2\sqrt{5}+3\sqrt{5}+4\sqrt{5}\) बनते हैं। कुल \(10\sqrt{5}\) नहीं बल्कि \(10\sqrt{5}\) है, विकल्पों को ध्यान से जाँचें।
Both sides are positive and (\(\sqrt{2}+\sqrt{3}\)2=5+2\sqrt{6}>5). So the first side is larger.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}+\sqrt{3}>\sqrt{5}\). Both sides are positive and (\(\sqrt{2}+\sqrt{3}\)2=5+2\sqrt{6}>5). So the first side is larger.
Step 3
Exam Tip
दोनों पक्ष धनात्मक हैं और (\(\sqrt{2}+\sqrt{3}\)2=5+2\sqrt{6}>5) है। इसलिए पहला पक्ष बड़ा है।
\(\sqrt{8}=2\sqrt{2}\), so \(\sqrt{2}+\sqrt{8}=3\sqrt{2}\). The unlike root \(\sqrt{5}\) remains separate.
Step 2
Why this answer is correct
The correct answer is A. \(3\sqrt{2}+\sqrt{5}\). \(\sqrt{8}=2\sqrt{2}\), so \(\sqrt{2}+\sqrt{8}=3\sqrt{2}\). The unlike root \(\sqrt{5}\) remains separate.
Step 3
Exam Tip
\(\sqrt{8}=2\sqrt{2}\) इसलिए \(\sqrt{2}+\sqrt{8}=3\sqrt{2}\) होता है। असमान जड़ \(\sqrt{5}\) अलग रहती है।
Multiplying by the conjugate gives denominator (5-3=2) and numerator \(8+2\sqrt{15}\). The simplified form is \(4+\sqrt{15}\).
Step 2
Why this answer is correct
The correct answer is A. \(4+\sqrt{15}\). Multiplying by the conjugate gives denominator (5-3=2) and numerator \(8+2\sqrt{15}\). The simplified form is \(4+\sqrt{15}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (5-3=2) और अंश \(8+2\sqrt{15}\) बनता है। सरल रूप \(4+\sqrt{15}\) है।
Multiplying by the conjugate makes the denominator (1). The numerator is (\(\sqrt{3}+\sqrt{2}\)2=5+2\sqrt{6}).
Step 2
Why this answer is correct
The correct answer is A. \(5+2\sqrt{6}\). Multiplying by the conjugate makes the denominator (1). The numerator is (\(\sqrt{3}+\sqrt{2}\)2=5+2\sqrt{6}).
Step 3
Exam Tip
हर के संयुग्मी से गुणा करने पर हर (1) बनता है। अंश (\(\sqrt{3}+\sqrt{2}\)2=5+2\sqrt{6}) है।
\(\sqrt{8}=2\sqrt{2}\), so the sum is \(3\sqrt{2}\). A non zero rational multiple of \(\sqrt{2}\) remains irrational.
Step 2
Why this answer is correct
The correct answer is A. यह \(3\sqrt{2}\) है / It is \(3\sqrt{2}\). \(\sqrt{8}=2\sqrt{2}\), so the sum is \(3\sqrt{2}\). A non zero rational multiple of \(\sqrt{2}\) remains irrational.
Step 3
Exam Tip
\(\sqrt{8}=2\sqrt{2}\), इसलिए योग \(3\sqrt{2}\) है। गैर शून्य परिमेय गुणक के साथ \(\sqrt{2}\) अपरिमेय रहता है।
Actually \(x=\sqrt{2}+\sqrt{3}\) satisfies \(x^4-10x^2+1=0\), not a simple quadratic here. Read powers carefully in such trick questions.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-10x^0+1=0\). Actually \(x=\sqrt{2}+\sqrt{3}\) satisfies \(x^4-10x^2+1=0\), not a simple quadratic here. Read powers carefully in such trick questions.
Step 3
Exam Tip
\(x^2=5+2\sqrt{6}\) और संयुग्मी के साथ गुणन से \(x^4-10x^2+1=0\) मिलता है। दिए विकल्प में \(x^0=1\) इसलिए पहला रूप सही नहीं दिखता, ध्यान से पढ़ें।
Multiplying by the conjugate makes the denominator (5-2=3). So the rationalized form is \(\frac{\sqrt{5}-\sqrt{2}}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\sqrt{5}-\sqrt{2}}{3}\). Multiplying by the conjugate makes the denominator (5-2=3). So the rationalized form is \(\frac{\sqrt{5}-\sqrt{2}}{3}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (5-2=3) हो जाता है। इसलिए परिमेय हर वाला रूप \(\frac{\sqrt{5}-\sqrt{2}}{3}\) है।
The expression becomes \(3\sqrt{5}+4\sqrt{5}-5\sqrt{5}=2\sqrt{5}\). \(2\sqrt{5}\) is irrational.
Step 2
Why this answer is correct
The correct answer is A. अपरिमेय संख्या / Irrational number. The expression becomes \(3\sqrt{5}+4\sqrt{5}-5\sqrt{5}=2\sqrt{5}\). \(2\sqrt{5}\) is irrational.
Step 3
Exam Tip
अभिव्यक्ति \(3\sqrt{5}+4\sqrt{5}-5\sqrt{5}=2\sqrt{5}\) बनती है। \(2\sqrt{5}\) अपरिमेय है।
\(\sqrt{8}=2\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\), so \(a=5\sqrt{2}\). Its square is (50), a rational number.
Step 2
Why this answer is correct
The correct answer is A. परिमेय संख्या / Rational number. \(\sqrt{8}=2\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\), so \(a=5\sqrt{2}\). Its square is (50), a rational number.
Step 3
Exam Tip
\(\sqrt{8}=2\sqrt{2}\) और \(\sqrt{18}=3\sqrt{2}\), इसलिए \(a=5\sqrt{2}\)। इसका वर्ग (50) परिमेय है।
