If the sum of roots is (11) and product is (30), the equation is \(x^2-11x+30=0\). Remember the monic form formula.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-11x+30=0\). If the sum of roots is (11) and product is (30), the equation is \(x^2-11x+30=0\). Remember the monic form formula.
Step 3
Exam Tip
यदि मूलों का योग (11) और गुणनफल (30) है, तो समीकरण \(x^2-11x+30=0\) होगा। मोनिक रूप का सूत्र याद रखें।
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।
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 sum is \(4\sqrt{3}\) and the product is (\(2\sqrt{3}\)2-1=11). (S) equals the sum and (P) equals the product.
Step 2
Why this answer is correct
The correct answer is A. \(S=4\sqrt{3}\), (P=11). The sum is \(4\sqrt{3}\) and the product is (\(2\sqrt{3}\)2-1=11). (S) equals the sum and (P) equals the product.
Step 3
Exam Tip
योग \(4\sqrt{3}\) और गुणनफल (\(2\sqrt{3}\)2-1=11) है। (S) योग और (P) गुणनफल के बराबर है।
A. योग \(6\sqrt{2}\), गुणनफल (17)/Sum \(6\sqrt{2}\), product (17)
Step 1
Concept
In a monic quadratic, the sum is (-b) and the product is (c). Therefore the sum is \(6\sqrt{2}\) and the product is (17).
Step 2
Why this answer is correct
The correct answer is A. योग \(6\sqrt{2}\), गुणनफल (17) / Sum \(6\sqrt{2}\), product (17). In a monic quadratic, the sum is (-b) and the product is (c). Therefore the sum is \(6\sqrt{2}\) and the product is (17).
Step 3
Exam Tip
एकक द्विघात में योग (-b) और गुणनफल (c) होता है। इसलिए योग \(6\sqrt{2}\) और गुणनफल (17) 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=\(\alpha+\beta\)2-4\alpha\beta=32-4(-2)=17). This method gives the answer without finding the zeroes.
Step 2
Why this answer is correct
The correct answer is A. (17). (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=32-4(-2)=17). This method gives the answer without finding the zeroes.
Step 3
Exam Tip
(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=32-4(-2)=17)। यह तरीका शून्यक निकाले बिना उत्तर देता है।
\(\alpha+\beta=2\) and \(\alpha\beta=-4\), so (\alpha-2+\beta-2=22-2(-4)=12). Symmetric values can be found without finding the zeroes.
Step 2
Why this answer is correct
The correct answer is A. (12). \(\alpha+\beta=2\) and \(\alpha\beta=-4\), so (\alpha-2+\beta-2=22-2(-4)=12). Symmetric values can be found without finding the zeroes.
Step 3
Exam Tip
\(\alpha+\beta=2\) और \(\alpha\beta=-4\), इसलिए (\alpha-2+\beta-2=22-2(-4)=12)। शून्यक निकाले बिना सममित मान निकाल सकते हैं।
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. गुणनफल (5), योग \(2\sqrt{6}\)/Product (5), sum \(2\sqrt{6}\)
Step 1
Concept
In a monic quadratic, the sum is (-b) and the product is (c). Here the sum is \(2\sqrt{6}\) and the product is (5).
Step 2
Why this answer is correct
The correct answer is A. गुणनफल (5), योग \(2\sqrt{6}\) / Product (5), sum \(2\sqrt{6}\). In a monic quadratic, the sum is (-b) and the product is (c). Here the sum is \(2\sqrt{6}\) and the product is (5).
Step 3
Exam Tip
एकक द्विघात में योग (-b) और गुणनफल (c) होता है। यहाँ योग \(2\sqrt{6}\) और गुणनफल (5) 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।
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।
A. \(\sqrt{2}\) और \(\sqrt{3}\)/\(\sqrt{2}\) and \(\sqrt{3}\)
Step 1
Concept
The sum \(\sqrt{2}+\sqrt{3}\) and product \(\sqrt{6}\) match the option \(\sqrt{2}\), \(\sqrt{3}\). Hence those are the zeroes.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\) और \(\sqrt{3}\) / \(\sqrt{2}\) and \(\sqrt{3}\). The sum \(\sqrt{2}+\sqrt{3}\) and product \(\sqrt{6}\) match the option \(\sqrt{2}\), \(\sqrt{3}\). Hence those are the zeroes.
Step 3
Exam Tip
योग \(\sqrt{2}+\sqrt{3}\) और गुणनफल \(\sqrt{6}\) विकल्प \(\sqrt{2}\), \(\sqrt{3}\) से मिलते हैं। इसलिए वही शून्यक हैं।
The radical terms cancel in the sum and the product is (9-8=1). This method is quick for conjugates.
Step 2
Why this answer is correct
The correct answer is A. योग (6), गुणनफल (1) / Sum (6), product (1). The radical terms cancel in the sum and the product is (9-8=1). This method is quick for conjugates.
Step 3
Exam Tip
योग में जड़ वाले पद कटते हैं और गुणनफल (9-8=1) है। संयुग्मी संख्याओं में यह तरीका तेज है।
The radical terms cancel in the sum and the product is (4-3=1). This method is quick for conjugates.
Step 2
Why this answer is correct
The correct answer is A. योग (4), गुणनफल (1) / Sum (4), product (1). The radical terms cancel in the sum and the product is (4-3=1). This method is quick for conjugates.
Step 3
Exam Tip
योग में जड़ वाले पद कटते हैं और गुणनफल (4-3=1) है। संयुग्मी संख्याओं में यह तरीका तेज है।
