irrational-coefficient se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
16 questions tagged with irrational-coefficient.
Which statement is correct for \(x^2+\sqrt{3}x+1\)?
A. यह बहुपद हैIt is a polynomial
\(\sqrt{3}\) वास्तविक गुणांक है और (x) की घातें पूर्ण संख्याएँ हैं। इसलिए यह बहुपद है। / \(\sqrt{3}\) is a real coefficient and the powers of (x) are whole numbers. Therefore it is a polynomial.
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Which expression is a polynomial in one variable with real coefficients?
A. \(\sqrt{2}x^2-3x+1\)
\(\sqrt{2}\) एक वास्तविक संख्या है और (x) की घातें पूर्ण संख्याएँ हैं। इसलिए यह बहुपद है। / \(\sqrt{2}\) is a real number and the powers of (x) are whole numbers. So it is a polynomial.
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If (p) and (q) are rational numbers and \(p+q\sqrt{5}\) is rational, what is true about (q)?
A. (q=0)
यदि \(q\ne0\), तो \(q\sqrt{5}\) अपरिमेय होगा और योग परिमेय नहीं हो सकता। परीक्षा में शून्य गुणांक की संभावना देखें। / If \(q\ne0\), then \(q\sqrt{5}\) is irrational and the sum cannot be rational. In exams check the possibility of a zero coefficient.
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If (p(x)=x-2-6\sqrt{2}x+17), what is the difference between its zeroes?
A. (2)
(D=\(6\sqrt{2}\)2-68=72-68=4), इसलिए शून्यक \(3\sqrt{2}\pm1\) हैं। उनका अंतर (2) है। / (D=\(6\sqrt{2}\)2-68=72-68=4), so the zeroes are \(3\sqrt{2}\pm1\). Their difference is (2).
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If (p(x)=x-2-6\sqrt{2}x+17), what are the sum and product of its zeroes?
A. योग \(6\sqrt{2}\), गुणनफल (17)Sum \(6\sqrt{2}\), product (17)
एकक द्विघात में योग (-b) और गुणनफल (c) होता है। इसलिए योग \(6\sqrt{2}\) और गुणनफल (17) है। / 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).
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If (p(x)=x-2-2\sqrt{2}x+1), what is the type of its zeroes?
A. दो भिन्न वास्तविक अपरिमेयTwo distinct real irrational
(D=\(2\sqrt{2}\)2-4=8-4=4) है और शून्यक \(\sqrt{2}\pm1\) हैं। ये वास्तविक और अपरिमेय हैं। / (D=\(2\sqrt{2}\)2-4=8-4=4), and the zeroes are \(\sqrt{2}\pm1\). They are real and irrational.
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If (p(x)=x-2+2\sqrt{7}x+6), what are its zeroes?
A. \(-\sqrt{7}+1\) और \(-\sqrt{7}-1\)\(-\sqrt{7}+1\) and \(-\sqrt{7}-1\)
सूत्र से \(x=\frac{-2\sqrt{7}\pm2}{2}=-\sqrt{7}\pm1\)। पहले विविक्तकर सरल करने से उत्तर साफ मिलता है। / Using the formula, \(x=\frac{-2\sqrt{7}\pm2}{2}=-\sqrt{7}\pm1\). Simplifying the discriminant first gives a clean answer.
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If (p(x)=x-2+2\sqrt{7}x+6), what is the discriminant?
A. (4)
(D=\(2\sqrt{7}\)2-4\cdot1\cdot6=28-24=4)। अपरिमेय गुणांक होने पर भी विविक्तकर परिमेय हो सकता है। / (D=\(2\sqrt{7}\)2-4\cdot1\cdot6=28-24=4). Even with an irrational coefficient, the discriminant can be rational.
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Which option correctly gives the zeroes of \(x^2-2\sqrt{3}x-1\)?
A. \(\sqrt{3}\pm2\)
सूत्र से \(x=\frac{2\sqrt{3}\pm\sqrt{12+4}}{2}=\sqrt{3}\pm2\)। \(\sqrt{16}=4\) को ध्यान से सरल करें। / Using the formula, \(x=\frac{2\sqrt{3}\pm\sqrt{12+4}}{2}=\sqrt{3}\pm2\). Simplify \(\sqrt{16}=4\) carefully.
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If (p(x)=x-2-2\sqrt{6}x+5), what are the product and sum of its zeroes?
A. गुणनफल (5), योग \(2\sqrt{6}\)Product (5), sum \(2\sqrt{6}\)
एकक द्विघात में योग (-b) और गुणनफल (c) होता है। यहाँ योग \(2\sqrt{6}\) और गुणनफल (5) है। / 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).
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If (p(x)=x-2-\(1+\sqrt{3}\)x+\sqrt{3}), what are its zeroes?
A. (1) और \(\sqrt{3}\)(1) and \(\sqrt{3}\)
शून्यकों का योग \(1+\sqrt{3}\) और गुणनफल \(\sqrt{3}\) है। (1) और \(\sqrt{3}\) दोनों शर्तें पूरी करते हैं। / The sum of zeroes is \(1+\sqrt{3}\) and the product is \(\sqrt{3}\). The numbers (1) and \(\sqrt{3}\) satisfy both conditions.
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If (p(x)=\sqrt{3}x-2-6x+2\sqrt{3}), what is the product of its zeroes?
A. (2)
गुणनफल \(\frac{c}{a}=\frac{2\sqrt{3}}{\sqrt{3}}=2\) है। समान करणी कट सकती है। / The product is \(\frac{c}{a}=\frac{2\sqrt{3}}{\sqrt{3}}=2\). Like radicals can cancel.
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If (p(x)=\sqrt{2}x-2-4x+\sqrt{2}), what is the sum of its zeroes?
A. \(2\sqrt{2}\)
योग \(-\frac{b}{a}=\frac{4}{\sqrt{2}}=2\sqrt{2}\) है। हर का परिमेयकरण करने से उत्तर सरल होता है। / The sum is \(-\frac{b}{a}=\frac{4}{\sqrt{2}}=2\sqrt{2}\). Rationalising the denominator simplifies the answer.
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If (p(x)=x-2-4\sqrt{2}x+8), in which form can it be written?
A. (\(x-2\sqrt{2}\)2)
क्योंकि (\(2\sqrt{2}\)2=8) और बीच का पद \(-4\sqrt{2}x\) है। अतः यह पूर्ण वर्ग है। / Because (\(2\sqrt{2}\)2=8) and the middle term is \(-4\sqrt{2}x\). Hence it is a perfect square.
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If (p(x)=2x-2-4\sqrt{2}x+4), what is the sum of its zeroes?
A. \(2\sqrt{2}\)
योग \(-\frac{b}{a}=-\frac{-4\sqrt{2}}{2}=2\sqrt{2}\) है। गुणांक देखते समय (a) को मत भूलें। / The sum is \(-\frac{b}{a}=-\frac{-4\sqrt{2}}{2}=2\sqrt{2}\). Do not forget the coefficient (a).
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If (p(x)=x-2-2\sqrt{3}x+2), what is the product of its zeroes?
A. (2)
द्विघात \(ax^2+bx+c\) में शून्यकों का गुणनफल \(\frac{c}{a}\) होता है। यहाँ \(\frac{2}{1}=2\) है। / In \(ax^2+bx+c\), the product of zeroes is \(\frac{c}{a}\). Here it is \(\frac{2}{1}=2\).
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