A. \(9+\sqrt{7}\) और \(9-\sqrt{7}\)/\(9+\sqrt{7}\) and \(9-\sqrt{7}\)
Step 1
Concept
The sum of \(9+\sqrt{7}\) and \(9-\sqrt{7}\) is (18), and the product is (81-7=74). In exams check both sum and product of options.
Step 2
Why this answer is correct
The correct answer is A. \(9+\sqrt{7}\) और \(9-\sqrt{7}\) / \(9+\sqrt{7}\) and \(9-\sqrt{7}\). The sum of \(9+\sqrt{7}\) and \(9-\sqrt{7}\) is (18), and the product is (81-7=74). In exams check both sum and product of options.
Step 3
Exam Tip
\(9+\sqrt{7}\) और \(9-\sqrt{7}\) का योग (18) और गुणनफल (81-7=74) है। परीक्षा में विकल्पों का योग और गुणनफल दोनों जांचें।
B. \(5+\sqrt{5}\) और \(5-\sqrt{5}\)/\(5+\sqrt{5}\) and \(5-\sqrt{5}\)
Step 1
Concept
The pair \(5+\sqrt{5}\) and \(5-\sqrt{5}\) has sum (10) and product (20) so it also fails. The pair (5+2) and (5-2) would be rational so none of the given options fits.
Step 2
Why this answer is correct
The correct answer is B. \(5+\sqrt{5}\) और \(5-\sqrt{5}\) / \(5+\sqrt{5}\) and \(5-\sqrt{5}\). The pair \(5+\sqrt{5}\) and \(5-\sqrt{5}\) has sum (10) and product (20) so it also fails. The pair (5+2) and (5-2) would be rational so none of the given options fits.
Step 3
Exam Tip
\(5+\sqrt{5}\) और \(5-\sqrt{5}\) का योग (10) और गुणनफल (25-5=20) है इसलिए यह भी नहीं है। सही युग्म (5+2) और (5-2) परिमेय होगा इसलिए दिए विकल्पों में कोई नहीं।
\(\alpha+\beta=14\) and \(\alpha\beta=43\) so (\alpha-2+\beta-2=(14)2-2(43)=110). In exams use (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta).
Step 2
Why this answer is correct
The correct answer is A. (110). \(\alpha+\beta=14\) and \(\alpha\beta=43\) so (\alpha-2+\beta-2=(14)2-2(43)=110). In exams use (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta).
Step 3
Exam Tip
\(\alpha+\beta=14\) और \(\alpha\beta=43\) है इसलिए (\alpha-2+\beta-2=(14)2-2(43)=110)। परीक्षा में पहचान (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta) लगाएं।
The sum is (12) and the product is (36-5=31). In exams find the sum and product of conjugate pairs separately.
Step 2
Why this answer is correct
The correct answer is A. (12) और (31) / (12) and (31). The sum is (12) and the product is (36-5=31). In exams find the sum and product of conjugate pairs separately.
Step 3
Exam Tip
योग (12) और गुणनफल (36-5=31) है। परीक्षा में संयुग्मी जोड़े का योग और गुणनफल अलग निकालें।
The product (\(3+\sqrt{2}\)\(3-\sqrt{2}\)=9-2=7), so (k=7). In exams connect the constant term with the product of zeroes.
Step 2
Why this answer is correct
The correct answer is A. (7). The product (\(3+\sqrt{2}\)\(3-\sqrt{2}\)=9-2=7), so (k=7). In exams connect the constant term with the product of zeroes.
Step 3
Exam Tip
गुणनफल (\(3+\sqrt{2}\)\(3-\sqrt{2}\)=9-2=7) है, इसलिए (k=7) होगा। परीक्षा में स्थिर पद को शून्यकों के गुणनफल से जोड़ें।
From the product \(k^2-7=9\), we get \(k^2=16\), and (k=4) fits the given form. In exams use the product to find the unknown.
Step 2
Why this answer is correct
The correct answer is A. (4). From the product \(k^2-7=9\), we get \(k^2=16\), and (k=4) fits the given form. In exams use the product to find the unknown.
Step 3
Exam Tip
गुणनफल \(k^2-7=9\) से \(k^2=16\) मिलता है और दिए रूप में (k=4) उपयुक्त है। परीक्षा में गुणनफल से अज्ञात निकालें।
A. \(4+\sqrt{5}\) और \(4-\sqrt{5}\)/\(4+\sqrt{5}\) and \(4-\sqrt{5}\)
Step 1
Concept
The sum of \(4+\sqrt{5}\) and \(4-\sqrt{5}\) is (8), and the product is (16-5=11). In exams check the sum and product of options.
Step 2
Why this answer is correct
The correct answer is A. \(4+\sqrt{5}\) और \(4-\sqrt{5}\) / \(4+\sqrt{5}\) and \(4-\sqrt{5}\). The sum of \(4+\sqrt{5}\) and \(4-\sqrt{5}\) is (8), and the product is (16-5=11). In exams check the sum and product of options.
Step 3
Exam Tip
\(4+\sqrt{5}\) और \(4-\sqrt{5}\) का योग (8) और गुणनफल (16-5=11) है। परीक्षा में विकल्पों का योग और गुणनफल जांचें।
(p(x)=x\(x-\sqrt{5}\)), so the zeroes are (0) and \(\sqrt{5}\). Taking the common factor is a fast method in exams.
Step 2
Why this answer is correct
The correct answer is A. \(0,\sqrt{5}\). (p(x)=x\(x-\sqrt{5}\)), so the zeroes are (0) and \(\sqrt{5}\). Taking the common factor is a fast method in exams.
Step 3
Exam Tip
(p(x)=x\(x-\sqrt{5}\)), इसलिए शून्यक (0) और \(\sqrt{5}\) हैं। परीक्षा में सामान्य गुणनखंड निकालना तेज तरीका है।
The sum is \(1+\sqrt{3}\) and the product is \(\sqrt{3}\), so the zeroes are (1) and \(\sqrt{3}\). Compare with \(x^2-Sx+P\) in exams.
Step 2
Why this answer is correct
The correct answer is A. \(1,\sqrt{3}\). The sum is \(1+\sqrt{3}\) and the product is \(\sqrt{3}\), so the zeroes are (1) and \(\sqrt{3}\). Compare with \(x^2-Sx+P\) in exams.
Step 3
Exam Tip
योग \(1+\sqrt{3}\) और गुणनफल \(\sqrt{3}\) है, इसलिए शून्यक (1) और \(\sqrt{3}\) हैं। परीक्षा में \(x^2-Sx+P\) से तुलना करें।
A. \(-\sqrt{7}+1\) और \(-\sqrt{7}-1\)/\(-\sqrt{7}+1\) and \(-\sqrt{7}-1\)
Step 1
Concept
Using the formula, \(x=\frac{-2\sqrt{7}\pm2}{2}=-\sqrt{7}\pm1\). Simplifying the discriminant first gives a clean answer.
Step 2
Why this answer is correct
The correct answer is A. \(-\sqrt{7}+1\) और \(-\sqrt{7}-1\) / \(-\sqrt{7}+1\) and \(-\sqrt{7}-1\). Using the formula, \(x=\frac{-2\sqrt{7}\pm2}{2}=-\sqrt{7}\pm1\). Simplifying the discriminant first gives a clean answer.
Step 3
Exam Tip
सूत्र से \(x=\frac{-2\sqrt{7}\pm2}{2}=-\sqrt{7}\pm1\)। पहले विविक्तकर सरल करने से उत्तर साफ मिलता है।
The zeroes are \(3\sqrt{2}\) and \(-3\sqrt{2}\). Therefore the sum is (0) and the product is (-18).
Step 2
Why this answer is correct
The correct answer is A. गुणनफल (-18), योग (0) / Product (-18), sum (0). The zeroes are \(3\sqrt{2}\) and \(-3\sqrt{2}\). Therefore the sum is (0) and the product is (-18).
Step 3
Exam Tip
शून्यक \(3\sqrt{2}\) और \(-3\sqrt{2}\) हैं। इसलिए योग (0) और गुणनफल (-18) है।
The sum is \(3+\sqrt{2}\) and the product is \(3\sqrt{2}\). These match (3) and \(\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. (3) और \(\sqrt{2}\) / (3) and \(\sqrt{2}\). The sum is \(3+\sqrt{2}\) and the product is \(3\sqrt{2}\). These match (3) and \(\sqrt{2}\).
Step 3
Exam Tip
योग \(3+\sqrt{2}\) और गुणनफल \(3\sqrt{2}\) है। ये (3) और \(\sqrt{2}\) से मिलते हैं।
\(\alpha+\beta=2\) and \(\alpha\beta=-11\), so (\alpha-2+\beta-2+\alpha\beta=\(\alpha+\beta\)2-\alpha\beta=4+11=15). Sum and product are enough for symmetric expressions.
Step 2
Why this answer is correct
The correct answer is A. (15). \(\alpha+\beta=2\) and \(\alpha\beta=-11\), so (\alpha-2+\beta-2+\alpha\beta=\(\alpha+\beta\)2-\alpha\beta=4+11=15). Sum and product are enough for symmetric expressions.
Step 3
Exam Tip
\(\alpha+\beta=2\) और \(\alpha\beta=-11\), इसलिए (\alpha-2+\beta-2+\alpha\beta=\(\alpha+\beta\)2-\alpha\beta=4+11=15)। सममित व्यंजकों में योग और गुणनफल काफी होते हैं।
The product is \(ab=\sqrt{2}\cdot\sqrt{18}=\sqrt{36}=6\). In radical multiplication, simplify the product inside the root first.
Step 2
Why this answer is correct
The correct answer is A. (6). The product is \(ab=\sqrt{2}\cdot\sqrt{18}=\sqrt{36}=6\). In radical multiplication, simplify the product inside the root first.
Step 3
Exam Tip
गुणनफल \(ab=\sqrt{2}\cdot\sqrt{18}=\sqrt{36}=6\) है। मूलों के गुणन में पहले अंदर के गुणनफल को सरल करें।
Using the formula, \(x=\frac{2\sqrt{3}\pm\sqrt{12+4}}{2}=\sqrt{3}\pm2\). Simplify \(\sqrt{16}=4\) carefully.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{3}\pm2\). Using the formula, \(x=\frac{2\sqrt{3}\pm\sqrt{12+4}}{2}=\sqrt{3}\pm2\). Simplify \(\sqrt{16}=4\) carefully.
Step 3
Exam Tip
सूत्र से \(x=\frac{2\sqrt{3}\pm\sqrt{12+4}}{2}=\sqrt{3}\pm2\)। \(\sqrt{16}=4\) को ध्यान से सरल करें।
By the formula, \(x=\frac{8\pm\sqrt{64-8}}{4}=2\pm\frac{\sqrt{14}}{2}\). Divide the whole numerator by the denominator carefully.
Step 2
Why this answer is correct
The correct answer is A. \(2\pm\frac{\sqrt{14}}{2}\). By the formula, \(x=\frac{8\pm\sqrt{64-8}}{4}=2\pm\frac{\sqrt{14}}{2}\). Divide the whole numerator by the denominator carefully.
Step 3
Exam Tip
सूत्र से \(x=\frac{8\pm\sqrt{64-8}}{4}=2\pm\frac{\sqrt{14}}{2}\) है। हर से भाग देते समय पूरे अंश को बाँटें।
\(\alpha+\beta=4\) and \(\alpha\beta=-1\), so \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{4}{-1}=-4\). Find sum and product first.
Step 2
Why this answer is correct
The correct answer is A. (-4). \(\alpha+\beta=4\) and \(\alpha\beta=-1\), so \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{4}{-1}=-4\). Find sum and product first.
Step 3
Exam Tip
\(\alpha+\beta=4\) और \(\alpha\beta=-1\), इसलिए \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{4}{-1}=-4\)। पहले योग और गुणनफल निकालें।
A. \(\sqrt{5}\) और \(\sqrt{7}\)/\(\sqrt{5}\) and \(\sqrt{7}\)
Step 1
Concept
The sum is \(\sqrt{5}+\sqrt{7}\) and the product is \(\sqrt{35}\). Both match \(\sqrt{5}\) and \(\sqrt{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{5}\) और \(\sqrt{7}\) / \(\sqrt{5}\) and \(\sqrt{7}\). The sum is \(\sqrt{5}+\sqrt{7}\) and the product is \(\sqrt{35}\). Both match \(\sqrt{5}\) and \(\sqrt{7}\).
Step 3
Exam Tip
योग \(\sqrt{5}+\sqrt{7}\) और गुणनफल \(\sqrt{35}\) है। ये दोनों \(\sqrt{5}\) और \(\sqrt{7}\) से मिलते हैं।
A. (p(x)) के शून्यक परिमेय हैं और (q(x)) के अपरिमेय वास्तविक हैं/Zeroes of (p(x)) are rational and zeroes of (q(x)) are irrational real
Step 1
Concept
For (p(x)), (D=196-180=16), while for (q(x)), (D=196-160=36), so both are rational. Therefore the listed intended contrast is not valid.
Step 2
Why this answer is correct
The correct answer is A. (p(x)) के शून्यक परिमेय हैं और (q(x)) के अपरिमेय वास्तविक हैं / Zeroes of (p(x)) are rational and zeroes of (q(x)) are irrational real. For (p(x)), (D=196-180=16), while for (q(x)), (D=196-160=36), so both are rational. Therefore the listed intended contrast is not valid.
Step 3
Exam Tip
(p(x)) के लिए (D=196-180=16), जबकि (q(x)) के लिए (D=196-160=36) है, इसलिए दोनों परिमेय हैं। इसलिए सही कथन विकल्प में नहीं है।
A. \(\sqrt{2}\) और \(\sqrt{3}\)/\(\sqrt{2}\) and \(\sqrt{3}\)
Step 1
Concept
The sum is \(\sqrt{2}+\sqrt{3}\) and the product is \(\sqrt{6}\). These match \(\sqrt{2}\) and \(\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\) और \(\sqrt{3}\) / \(\sqrt{2}\) and \(\sqrt{3}\). The sum is \(\sqrt{2}+\sqrt{3}\) and the product is \(\sqrt{6}\). These match \(\sqrt{2}\) and \(\sqrt{3}\).
Step 3
Exam Tip
योग \(\sqrt{2}+\sqrt{3}\) और गुणनफल \(\sqrt{6}\) है। ये \(\sqrt{2}\) और \(\sqrt{3}\) से मिलते हैं।
From \(5x^2-5=0\), \(x^2=1\), so \(x=\pm1\). Do not mistakenly take \(\sqrt{5}\) because of the common factor.
Step 2
Why this answer is correct
The correct answer is A. (1,-1), परिमेय / (1,-1), rational. From \(5x^2-5=0\), \(x^2=1\), so \(x=\pm1\). Do not mistakenly take \(\sqrt{5}\) because of the common factor.
Step 3
Exam Tip
\(5x^2-5=0\) से \(x^2=1\), इसलिए \(x=\pm1\) हैं। सामान्य गुणनखंड से भ्रमित होकर \(\sqrt{5}\) न लें।
By the formula, \(x=\frac{4\pm\sqrt{16+24}}{2}=2\pm\sqrt{10}\). Remember \(\sqrt{40}=2\sqrt{10}\) while simplifying (D).
Step 2
Why this answer is correct
The correct answer is A. \(2\pm\sqrt{10}\). By the formula, \(x=\frac{4\pm\sqrt{16+24}}{2}=2\pm\sqrt{10}\). Remember \(\sqrt{40}=2\sqrt{10}\) while simplifying (D).
Step 3
Exam Tip
सूत्र से \(x=\frac{4\pm\sqrt{16+24}}{2}=2\pm\sqrt{10}\) है। (D) को सरल करने में \(\sqrt{40}=2\sqrt{10}\) याद रखें।
The sum of zeroes is \(1+\sqrt{3}\) and the product is \(\sqrt{3}\). The numbers (1) and \(\sqrt{3}\) satisfy both conditions.
Step 2
Why this answer is correct
The correct answer is A. (1) और \(\sqrt{3}\) / (1) and \(\sqrt{3}\). The sum of zeroes is \(1+\sqrt{3}\) and the product is \(\sqrt{3}\). The numbers (1) and \(\sqrt{3}\) satisfy both conditions.
Step 3
Exam Tip
शून्यकों का योग \(1+\sqrt{3}\) और गुणनफल \(\sqrt{3}\) है। (1) और \(\sqrt{3}\) दोनों शर्तें पूरी करते हैं।
Since (3x-2-12x+6=3\(x^2-4x+2\)), the zeroes are \(2\pm\sqrt{2}\). Removing a common factor first makes calculation easier.
Step 2
Why this answer is correct
The correct answer is A. \(2\pm\sqrt{2}\). Since (3x-2-12x+6=3\(x^2-4x+2\)), the zeroes are \(2\pm\sqrt{2}\). Removing a common factor first makes calculation easier.
Step 3
Exam Tip
(3x-2-12x+6=3\(x^2-4x+2\)), इसलिए शून्यक \(2\pm\sqrt{2}\) हैं। पहले सामान्य गुणनखंड हटाना गणना आसान करता है।
