C. सिर्फ \(\sqrt{2}\) ही अकेला अपरिमेय शून्यक हो और गुणांक परिमेय रहें/Only \(\sqrt{2}\) is the sole irrational zero while coefficients stay rational
Step 1
Concept
In a quadratic with rational coefficients an irrational zero comes with its conjugate. In exams be suspicious of a lone irrational root.
Step 2
Why this answer is correct
The correct answer is C. सिर्फ \(\sqrt{2}\) ही अकेला अपरिमेय शून्यक हो और गुणांक परिमेय रहें / Only \(\sqrt{2}\) is the sole irrational zero while coefficients stay rational. In a quadratic with rational coefficients an irrational zero comes with its conjugate. In exams be suspicious of a lone irrational root.
Step 3
Exam Tip
परिमेय गुणांकों वाले द्विघात में अपरिमेय शून्यक अपने संयुग्मी के साथ आता है। परीक्षा में अकेले अपरिमेय मूल पर संदेह करें।
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।
C. कोई वास्तविक मूल नहीं है/There are no real roots
Step 1
Concept
The discriminant is (4-20=-16), which is negative, so there are no real roots. In exams do not treat a negative discriminant as real zeroes.
Step 2
Why this answer is correct
The correct answer is C. कोई वास्तविक मूल नहीं है / There are no real roots. The discriminant is (4-20=-16), which is negative, so there are no real roots. In exams do not treat a negative discriminant as real zeroes.
Step 3
Exam Tip
विविक्तकर (4-20=-16) ऋणात्मक है, इसलिए वास्तविक मूल नहीं हैं। परीक्षा में ऋणात्मक विविक्तकर को वास्तविक शून्यक नहीं मानें।
C. दो अलग अपरिमेय वास्तविक/Two distinct irrational real values
Step 1
Concept
The discriminant is (36-4=32), positive but not a perfect square. So there are two distinct irrational real roots.
Step 2
Why this answer is correct
The correct answer is C. दो अलग अपरिमेय वास्तविक / Two distinct irrational real values. The discriminant is (36-4=32), positive but not a perfect square. So there are two distinct irrational real roots.
Step 3
Exam Tip
विविक्तकर (36-4=32) है, जो पूर्ण वर्ग नहीं है और धनात्मक है। इसलिए दो अलग अपरिमेय वास्तविक मूल मिलते हैं।
The sum is (4) and the product is (4-6=-2), so the polynomial is \(x^2-4x-2\). Remember the formula \(x^2-Sx+P\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4x-2\). The sum is (4) and the product is (4-6=-2), so the polynomial is \(x^2-4x-2\). Remember the formula \(x^2-Sx+P\).
Step 3
Exam Tip
योग (4) और गुणनफल (4-6=-2) है, इसलिए बहुपद \(x^2-4x-2\) है। परीक्षा में \(x^2-Sx+P\) सूत्र याद रखें।
\(\alpha+\beta=2\) 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।
\(\alpha+\beta=2\) 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।
\(\alpha+\beta=4\) 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।
The zeroes are \(5\pm2\sqrt{2}\), so the difference is \(4\sqrt{2}\). For conjugate zeroes, the difference is twice the radical part.
Step 2
Why this answer is correct
The correct answer is A. \(4\sqrt{2}\). The zeroes are \(5\pm2\sqrt{2}\), so the difference is \(4\sqrt{2}\). For conjugate zeroes, the difference is twice the radical part.
Step 3
Exam Tip
शून्यक \(5\pm2\sqrt{2}\) हैं, इसलिए अंतर \(4\sqrt{2}\) है। संयुग्मी शून्यकों में अंतर मूल भाग का दोगुना होता है।
The constant term is the product, and (\(4+\sqrt{11}\)\(4-\sqrt{11}\)=16-11=5). In conjugate products, the irrational middle part cancels.
Step 2
Why this answer is correct
The correct answer is A. (5). The constant term is the product, and (\(4+\sqrt{11}\)\(4-\sqrt{11}\)=16-11=5). In conjugate products, the irrational middle part cancels.
Step 3
Exam Tip
स्थिर पद गुणनफल है और (\(4+\sqrt{11}\)\(4-\sqrt{11}\)=16-11=5)। संयुग्मी गुणनफल में बीच का अपरिमेय भाग हट जाता है।
The other zero will be \(-\sqrt{7}\), so the sum is (0) and (a=-0=0). With rational coefficients, take the conjugate zero.
Step 2
Why this answer is correct
The correct answer is A. (0). The other zero will be \(-\sqrt{7}\), so the sum is (0) and (a=-0=0). With rational coefficients, take the conjugate zero.
Step 3
Exam Tip
दूसरा शून्यक \(-\sqrt{7}\) होगा, इसलिए योग (0) और (a=-0=0) है। परिमेय गुणांक में संयुग्मी शून्यक लेना जरूरी है।
With rational coefficients, the conjugate of an irrational zero is also a zero. So \(2-\sqrt{3}\) will be the other zero.
Step 2
Why this answer is correct
The correct answer is A. \(2-\sqrt{3}\). With rational coefficients, the conjugate of an irrational zero is also a zero. So \(2-\sqrt{3}\) will be the other zero.
Step 3
Exam Tip
परिमेय गुणांकों में अपरिमेय शून्यक का संयुग्मी भी शून्यक होता है। इसलिए \(2-\sqrt{3}\) दूसरा शून्यक होगा।
A. दो भिन्न अपरिमेय वास्तविक शून्यक/Two distinct irrational real zeroes
Step 1
Concept
The discriminant is (D=36-16=20), so the zeroes are \(3\pm\sqrt{5}\). If (D) is not a perfect square, real zeroes can be irrational.
Step 2
Why this answer is correct
The correct answer is A. दो भिन्न अपरिमेय वास्तविक शून्यक / Two distinct irrational real zeroes. The discriminant is (D=36-16=20), so the zeroes are \(3\pm\sqrt{5}\). If (D) is not a perfect square, real zeroes can be irrational.
Step 3
Exam Tip
विविक्तकर (D=36-16=20) है, इसलिए शून्यक \(3\pm\sqrt{5}\) हैं। (D) पूर्ण वर्ग न हो तो वास्तविक शून्यक अपरिमेय हो सकते हैं।
The sum of zeroes is (10) and the product is (19). Hence \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{10}{19}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{10}{19}\). The sum of zeroes is (10) and the product is (19). Hence \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{10}{19}\).
Step 3
Exam Tip
शून्यकों का योग (10) और गुणनफल (19) है। अतः \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{10}{19}\)।
The sum is \(2-\sqrt{5}\), so the coefficient of (x) is (-\(2-\sqrt{5}\)=\sqrt{5}-2). The product \(-2\sqrt{5}\) also matches.
Step 2
Why this answer is correct
The correct answer is A. (2) और \(-\sqrt{5}\) / (2) and \(-\sqrt{5}\). The sum is \(2-\sqrt{5}\), so the coefficient of (x) is (-\(2-\sqrt{5}\)=\sqrt{5}-2). The product \(-2\sqrt{5}\) also matches.
Step 3
Exam Tip
योग \(2-\sqrt{5}\) है, इसलिए (x) का गुणांक (-\(2-\sqrt{5}\)=\sqrt{5}-2) है। गुणनफल \(-2\sqrt{5}\) भी सही है।
C. दो भिन्न अपरिमेय शून्यक/Two distinct irrational zeroes
Step 1
Concept
The simplified form is \(x^2-4x+2\), and (D=16-8=8). Since (8) is not a perfect square, the zeroes are irrational.
Step 2
Why this answer is correct
The correct answer is C. दो भिन्न अपरिमेय शून्यक / Two distinct irrational zeroes. The simplified form is \(x^2-4x+2\), and (D=16-8=8). Since (8) is not a perfect square, the zeroes are irrational.
Step 3
Exam Tip
सरल रूप \(x^2-4x+2\) है और (D=16-8=8) है। (8) पूर्ण वर्ग नहीं है, इसलिए शून्यक अपरिमेय हैं।
C. दो भिन्न अपरिमेय शून्यक/Two distinct irrational zeroes
Step 1
Concept
The discriminant is (D=36-16=20), and (20) is not a perfect square. So the zeroes are real, distinct, and irrational.
Step 2
Why this answer is correct
The correct answer is C. दो भिन्न अपरिमेय शून्यक / Two distinct irrational zeroes. The discriminant is (D=36-16=20), and (20) is not a perfect square. So the zeroes are real, distinct, and irrational.
Step 3
Exam Tip
विविक्तकर (D=36-16=20) है और (20) पूर्ण वर्ग नहीं है। इसलिए शून्यक वास्तविक भिन्न और अपरिमेय हैं।
A. यह (x)-अक्ष को नहीं मिलेगा/It will not meet the (x)-axis
Step 1
Concept
(x-2+6x+10=(x+3)2+1), which is always positive. So there is no real zero.
Step 2
Why this answer is correct
The correct answer is A. यह (x)-अक्ष को नहीं मिलेगा / It will not meet the (x)-axis. (x-2+6x+10=(x+3)2+1), which is always positive. So there is no real zero.
Step 3
Exam Tip
(x-2+6x+10=(x+3)2+1) है जो हमेशा धनात्मक है। इसलिए कोई वास्तविक शून्यक नहीं है।
