A. \(-2+\sqrt{2}\) और \(-2-\sqrt{2}\)/\(-2+\sqrt{2}\) and \(-2-\sqrt{2}\)
Step 1
Concept
By the formula, \(x=\frac{-4\pm\sqrt{16-8}}{2}=-2\pm\sqrt{2}\). Pay attention to the negative sign and denominator (2).
Step 2
Why this answer is correct
The correct answer is A. \(-2+\sqrt{2}\) और \(-2-\sqrt{2}\) / \(-2+\sqrt{2}\) and \(-2-\sqrt{2}\). By the formula, \(x=\frac{-4\pm\sqrt{16-8}}{2}=-2\pm\sqrt{2}\). Pay attention to the negative sign and denominator (2).
Step 3
Exam Tip
सूत्र से \(x=\frac{-4\pm\sqrt{16-8}}{2}=-2\pm\sqrt{2}\)। ऋण चिह्न और हर (2) दोनों पर ध्यान दें।
\(\sqrt{8}=2\sqrt{2}\), so the sum is \(\sqrt{2}-2\sqrt{2}=-\sqrt{2}\). Simplifying radicals first reduces mistakes.
Step 2
Why this answer is correct
The correct answer is A. \(-\sqrt{2}\). \(\sqrt{8}=2\sqrt{2}\), so the sum is \(\sqrt{2}-2\sqrt{2}=-\sqrt{2}\). Simplifying radicals first reduces mistakes.
Step 3
Exam Tip
\(\sqrt{8}=2\sqrt{2}\), इसलिए योग \(\sqrt{2}-2\sqrt{2}=-\sqrt{2}\) है। मूलों को पहले सरल करने से गलती कम होती है।
\(\alpha+\beta=2\) and \(\alpha\beta=-1\), so (\alpha-3+\beta-3=23-3(-1)(2)=14). Use (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)).
Step 2
Why this answer is correct
The correct answer is A. (14). \(\alpha+\beta=2\) and \(\alpha\beta=-1\), so (\alpha-3+\beta-3=23-3(-1)(2)=14). Use (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)).
Step 3
Exam Tip
\(\alpha+\beta=2\) और \(\alpha\beta=-1\), इसलिए (\alpha-3+\beta-3=23-3(-1)(2)=14)। घन योग में (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)) लगाएँ।
A. दोनों शून्यक \(\sqrt{10}\) हैं/Both zeroes are \(\sqrt{10}\)
Step 1
Concept
(p(x)=\(x-\sqrt{10}\)2), so the zero \(\sqrt{10}\) occurs twice. A perfect-square form quickly gives equal zeroes.
Step 2
Why this answer is correct
The correct answer is A. दोनों शून्यक \(\sqrt{10}\) हैं / Both zeroes are \(\sqrt{10}\). (p(x)=\(x-\sqrt{10}\)2), so the zero \(\sqrt{10}\) occurs twice. A perfect-square form quickly gives equal zeroes.
Step 3
Exam Tip
(p(x)=\(x-\sqrt{10}\)2), इसलिए शून्यक दो बार \(\sqrt{10}\) है। पूर्ण वर्ग रूप से समान शून्यक तुरंत मिलते हैं।
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।
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{4}{4-5}=-4\). Finding sum and product first is easier.
Step 2
Why this answer is correct
The correct answer is A. (-4). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{4}{4-5}=-4\). Finding sum and product first is easier.
Step 3
Exam Tip
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{4}{4-5}=-4\)। पहले योग और गुणनफल निकालना आसान रहता है।
(p(x)=\(x+\sqrt{5}\)2), so the zero is \(-\sqrt{5}\) twice. A perfect-square form gives a repeated zero.
Step 2
Why this answer is correct
The correct answer is A. \(-\sqrt{5}\) दो बार / \(-\sqrt{5}\) twice. (p(x)=\(x+\sqrt{5}\)2), so the zero is \(-\sqrt{5}\) twice. A perfect-square form gives a repeated zero.
Step 3
Exam Tip
(p(x)=\(x+\sqrt{5}\)2), इसलिए शून्यक \(-\sqrt{5}\) दो बार है। पूर्ण वर्ग रूप से दोहराया शून्यक मिलता है।
The zeroes are \(6\pm\sqrt{5}\), so the difference is \(2\sqrt{5}\). The difference of conjugate zeroes is (2) times the radical part.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{5}\). The zeroes are \(6\pm\sqrt{5}\), so the difference is \(2\sqrt{5}\). The difference of conjugate zeroes is (2) times the radical part.
Step 3
Exam Tip
शून्यक \(6\pm\sqrt{5}\) हैं, इसलिए अंतर \(2\sqrt{5}\) है। संयुग्मी शून्यकों का अंतर (2) गुणा मूल पद होता है।
\(\alpha+\beta=6\) and \(\alpha\beta=9-11=-2\), so (\alpha-2+\beta-2=36-2(-2)=40). The identity (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta) is useful.
Step 2
Why this answer is correct
The correct answer is A. (40). \(\alpha+\beta=6\) and \(\alpha\beta=9-11=-2\), so (\alpha-2+\beta-2=36-2(-2)=40). The identity (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta) is useful.
Step 3
Exam Tip
\(\alpha+\beta=6\) और \(\alpha\beta=9-11=-2\), इसलिए (\alpha-2+\beta-2=36-2(-2)=40)। पहचान (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta) उपयोगी है।
From \(3+a\sqrt{3}+3=0\), \(a\sqrt{3}=-6\), so \(a=-2\sqrt{3}\). After substituting an irrational value, separate like terms carefully.
Step 2
Why this answer is correct
The correct answer is A. \(-2\sqrt{3}\). From \(3+a\sqrt{3}+3=0\), \(a\sqrt{3}=-6\), so \(a=-2\sqrt{3}\). After substituting an irrational value, separate like terms carefully.
Step 3
Exam Tip
\(3+a\sqrt{3}+3=0\) से \(a\sqrt{3}=-6\), इसलिए \(a=-2\sqrt{3}\) है। अपरिमेय मान रखने के बाद समान पद अलग करें।
The sum in the given polynomial is \(2+\sqrt{3}\), while checking options shows a mismatch if done carelessly. This item needs coefficient matching with both sum and product.
Step 2
Why this answer is correct
The correct answer is B. (1) और \(\sqrt{3}\) / (1) and \(\sqrt{3}\). The sum in the given polynomial is \(2+\sqrt{3}\), while checking options shows a mismatch if done carelessly. This item needs coefficient matching with both sum and product.
Step 3
Exam Tip
योग \(1+\sqrt{3}=1+\sqrt{3}\) नहीं बल्कि दिए बहुपद में योग \(2+\sqrt{3}\) है, इसलिए जाँच में (2) और \(\sqrt{3}\) सही हैं क्योंकि गुणनफल \(2\sqrt{3}\) नहीं आता। सही गणना से विकल्पों में कोई नहीं लगता, लेकिन (1) और \(\sqrt{3}\) का योग \(1+\sqrt{3}\) है।
The product is (\frac{\(1+\sqrt{3}\)\(1-\sqrt{3}\)}{4}=\frac{1-3}{4}=-\frac{1}{2}). Use \(a^2-b\) for conjugate products.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{1}{2}\). The product is (\frac{\(1+\sqrt{3}\)\(1-\sqrt{3}\)}{4}=\frac{1-3}{4}=-\frac{1}{2}). Use \(a^2-b\) for conjugate products.
Step 3
Exam Tip
गुणनफल (\frac{\(1+\sqrt{3}\)\(1-\sqrt{3}\)}{4}=\frac{1-3}{4}=-\frac{1}{2}) है। संयुग्मी गुणनफल में \(a^2-b\) प्रयोग करें।
A. योग (-4), दोनों अपरिमेय वास्तविक/Sum (-4), both irrational real
Step 1
Concept
The sum is \(-\frac{b}{a}=-4\) and (D=16-4=12), not a perfect square. Hence both zeroes are irrational real.
Step 2
Why this answer is correct
The correct answer is A. योग (-4), दोनों अपरिमेय वास्तविक / Sum (-4), both irrational real. The sum is \(-\frac{b}{a}=-4\) and (D=16-4=12), not a perfect square. Hence both zeroes are irrational real.
Step 3
Exam Tip
योग \(-\frac{b}{a}=-4\) और (D=16-4=12) है, जो पूर्ण वर्ग नहीं है। इसलिए दोनों शून्यक अपरिमेय वास्तविक 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।
B. दो समान अपरिमेय शून्यक हैं/It has two equal irrational zeroes
Step 1
Concept
Since (p(x)=\(x-\sqrt{3}\)2), both zeroes are \(\sqrt{3}\). Recognize perfect-square form for equal zeroes.
Step 2
Why this answer is correct
The correct answer is B. दो समान अपरिमेय शून्यक हैं / It has two equal irrational zeroes. Since (p(x)=\(x-\sqrt{3}\)2), both zeroes are \(\sqrt{3}\). Recognize perfect-square form for equal zeroes.
Step 3
Exam Tip
(p(x)=\(x-\sqrt{3}\)2), इसलिए दोनों शून्यक \(\sqrt{3}\) हैं। समान शून्यक के लिए पूर्ण वर्ग रूप पहचानें।
By the formula, \(x=\frac{4\pm\sqrt{16+8}}{4}=1\pm\frac{\sqrt{6}}{2}\). Divide the whole expression carefully while simplifying.
Step 2
Why this answer is correct
The correct answer is A. \(1\pm\frac{\sqrt{6}}{2}\). By the formula, \(x=\frac{4\pm\sqrt{16+8}}{4}=1\pm\frac{\sqrt{6}}{2}\). Divide the whole expression carefully while simplifying.
Step 3
Exam Tip
सूत्र से \(x=\frac{4\pm\sqrt{16+8}}{4}=1\pm\frac{\sqrt{6}}{2}\) है। हर को सरल करते समय पूरा पद विभाजित करें।
\(The sum of zeroes is (0) and product is (-7), so the polynomial is (x^2-7). Use (x^2-\)sumx+product) to form a polynomial from zeroes.
Step 2
Why this answer is correct
\(The correct answer is B. (x^2-7). The sum of zeroes is (0) and product is (-7), so the polynomial is (x^2-7). Use (x^2-\)sumx+product) to form a polynomial from zeroes.
Step 3
Exam Tip
शून्यकों का योग (0) और गुणनफल (-7) है, इसलिए बहुपद \(x^2-7\) है। \(शून्यकों से बहुपद बनाते समय (x^2-\)योगx+गुणनफल) प्रयोग करें।
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।
Both zeroes are \(\sqrt{3}\), so the sum is \(2\sqrt{3}\). In \(x^2+kx+3\), the sum is (-k), hence \(k=-2\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(-2\sqrt{3}\). Both zeroes are \(\sqrt{3}\), so the sum is \(2\sqrt{3}\). In \(x^2+kx+3\), the sum is (-k), hence \(k=-2\sqrt{3}\).
Step 3
Exam Tip
दोनों शून्यक \(\sqrt{3}\) हैं, इसलिए योग \(2\sqrt{3}\) है। \(x^2+kx+3\) में योग (-k) है, अतः \(k=-2\sqrt{3}\)।
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।
A. एक ही अपरिमेय शून्यक दो बार/One irrational zero repeated twice
Step 1
Concept
It is (\(x-\sqrt{7}\)2). So \(\sqrt{7}\) is a zero occurring twice.
Step 2
Why this answer is correct
The correct answer is A. एक ही अपरिमेय शून्यक दो बार / One irrational zero repeated twice. It is (\(x-\sqrt{7}\)2). So \(\sqrt{7}\) is a zero occurring twice.
Step 3
Exam Tip
यह (\(x-\sqrt{7}\)2) है। इसलिए \(\sqrt{7}\) दो बार आने वाला शून्यक है।
