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।
A. \(4+\sqrt{6}\) और \(4-\sqrt{6}\)/\(4+\sqrt{6}\) and \(4-\sqrt{6}\)
Step 1
Concept
For rational coefficients, the conjugate \(a-\sqrt{b}\) accompanies \(a+\sqrt{b}\). Hence the first pair is correct.
Step 2
Why this answer is correct
The correct answer is A. \(4+\sqrt{6}\) और \(4-\sqrt{6}\) / \(4+\sqrt{6}\) and \(4-\sqrt{6}\). For rational coefficients, the conjugate \(a-\sqrt{b}\) accompanies \(a+\sqrt{b}\). Hence the first pair is correct.
Step 3
Exam Tip
परिमेय गुणांकों के लिए \(a+\sqrt{b}\) का संयुग्मी \(a-\sqrt{b}\) साथ आता है। इसलिए पहला युग्म सही है।
The sum is \(3+\sqrt{5}+3-\sqrt{5}=6\), which is rational. Conjugate irrational numbers often have a rational sum.
Step 2
Why this answer is correct
The correct answer is B. परिमेय संख्या / Rational number. The sum is \(3+\sqrt{5}+3-\sqrt{5}=6\), which is rational. Conjugate irrational numbers often have a rational sum.
Step 3
Exam Tip
योग \(3+\sqrt{5}+3-\sqrt{5}=6\) है, जो परिमेय है। संयुग्मी अपरिमेय संख्याओं का योग अक्सर परिमेय होता है।
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here the sum is (10) and product is (25-6=19), so the answer is \(\frac{10}{19}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{10}{19}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here the sum is (10) and product is (25-6=19), so the answer is \(\frac{10}{19}\).
Step 3
Exam Tip
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\) होता है। यहाँ योग (10) और गुणनफल (25-6=19), इसलिए उत्तर \(\frac{10}{19}\) है।
A. दिए गए आधार पर कोई शून्यक नहीं दिखता/No zero is shown from the given data
Step 1
Concept
Zeroes are linked only with the (x)-axis where (y=0). Intersections with (y=2) do not show zeroes.
Step 2
Why this answer is correct
The correct answer is A. दिए गए आधार पर कोई शून्यक नहीं दिखता / No zero is shown from the given data. Zeroes are linked only with the (x)-axis where (y=0). Intersections with (y=2) do not show zeroes.
Step 3
Exam Tip
शून्यक केवल (x)-अक्ष यानी (y=0) से जुड़े होते हैं। (y=2) से प्रतिच्छेद शून्यक नहीं बताता।
A. इससे शून्यक निश्चित नहीं होता/A zero cannot be determined from this alone
Step 1
Concept
The (y)-intercept tells (p(0)) not all zeroes. Zeroes need (x)-axis intersections.
Step 2
Why this answer is correct
The correct answer is A. इससे शून्यक निश्चित नहीं होता / A zero cannot be determined from this alone. The (y)-intercept tells (p(0)) not all zeroes. Zeroes need (x)-axis intersections.
Step 3
Exam Tip
(y)-प्रतिच्छेद (p(0)) बताता है न कि सभी शून्यक। शून्यक के लिए (x)-अक्ष से प्रतिच्छेद चाहिए।
It is ((x-d)2-36), so \(x-d=\pm6\) and the zeroes are (d-6), (d+6). Tip: use difference of squares.
Step 2
Why this answer is correct
The correct answer is A. (d-6) और (d+6) / (d-6) and (d+6). It is ((x-d)2-36), so \(x-d=\pm6\) and the zeroes are (d-6), (d+6). Tip: use difference of squares.
Step 3
Exam Tip
यह ((x-d)2-36) है, इसलिए \(x-d=\pm6\) और शून्यक (d-6), (d+6) हैं। टिप: वर्गों के अंतर का उपयोग करें।
In this interval the signs are (+), (-), (-), so the product is positive. Tip: the product of two negative factors is positive.
Step 2
Why this answer is correct
The correct answer is A. (x)-अक्ष के ऊपर / Above the (x)-axis. In this interval the signs are (+), (-), (-), so the product is positive. Tip: the product of two negative factors is positive.
Step 3
Exam Tip
इस अंतराल में चिह्न (+), (-), (-) हैं, इसलिए गुणनफल धनात्मक है। टिप: दो ऋणात्मक कारकों का गुणन धनात्मक होता है।
A. (22) को (10) करना होगा/(22) must be changed to (10)
Step 1
Concept
For equal distance from the (y)-axis, zeroes should be opposites, so (10) is needed with (-10). Tip: symmetric zeroes are (a) and (-a).
Step 2
Why this answer is correct
The correct answer is A. (22) को (10) करना होगा / (22) must be changed to (10). For equal distance from the (y)-axis, zeroes should be opposites, so (10) is needed with (-10). Tip: symmetric zeroes are (a) and (-a).
Step 3
Exam Tip
(y)-अक्ष से समान दूरी के लिए शून्यक विपरीत होने चाहिए, इसलिए (-10) के साथ (10) चाहिए। टिप: सममित शून्यक (a) और (-a) होते हैं।
A. यह दोनों शून्यकों का मध्य है/It is the midpoint of the two zeroes
Step 1
Concept
\(18=\frac{10+26}{2}\), so it is the midpoint of the two zeroes. Tip: the midpoint value need not be a zero.
Step 2
Why this answer is correct
The correct answer is A. यह दोनों शून्यकों का मध्य है / It is the midpoint of the two zeroes. \(18=\frac{10+26}{2}\), so it is the midpoint of the two zeroes. Tip: the midpoint value need not be a zero.
Step 3
Exam Tip
\(18=\frac{10+26}{2}\), इसलिए यह दोनों शून्यकों का मध्य है। टिप: मध्य मान का शून्यक होना जरूरी नहीं।
The origin is also on the (x)-axis, and (x=-9) is another (x)-axis intersection. Tip: count ((0,0)) as zero (0).
Step 2
Why this answer is correct
The correct answer is A. (0) और (-9) / (0) and (-9). The origin is also on the (x)-axis, and (x=-9) is another (x)-axis intersection. Tip: count ((0,0)) as zero (0).
Step 3
Exam Tip
मूल बिंदु (x)-अक्ष पर भी है और (x=-9) भी (x)-अक्ष कटान है। टिप: ((0,0)) को शून्यक (0) के रूप में गिनें।
The other zero is (8), and the average is \(\frac{-10+8}{2}=-1\). Tip: the axis of symmetry is the average of two zeroes.
Step 2
Why this answer is correct
The correct answer is A. (x=-1). The other zero is (8), and the average is \(\frac{-10+8}{2}=-1\). Tip: the axis of symmetry is the average of two zeroes.
Step 3
Exam Tip
दूसरा शून्यक (8) है और औसत \(\frac{-10+8}{2}=-1\) है। टिप: सममिति अक्ष दो शून्यकों का औसत है।
The polynomial is ((x-n)(x-(n+3))), so the zeroes are (n) and (n+3). Tip: write zeroes as ((x,0)).
Step 2
Why this answer is correct
The correct answer is A. ((n,0)) और ((n+3,0)) / ((n,0)) and ((n+3,0)). The polynomial is ((x-n)(x-(n+3))), so the zeroes are (n) and (n+3). Tip: write zeroes as ((x,0)).
Step 3
Exam Tip
बहुपद ((x-n)(x-(n+3))) है इसलिए शून्यक (n) और (n+3) हैं। टिप: शून्यकों को ((x,0)) में लिखें।
(x=-4) lies between the two zeroes and an upward-opening parabola stays below there. Tip: check the sign region between zeroes.
Step 2
Why this answer is correct
The correct answer is A. (x)-अक्ष के नीचे / Below the (x)-axis. (x=-4) lies between the two zeroes and an upward-opening parabola stays below there. Tip: check the sign region between zeroes.
Step 3
Exam Tip
(x=-4) दोनों शून्यकों के बीच है और ऊपर खुलने वाला परवलय बीच में नीचे रहता है। टिप: शून्यकों के बीच संकेत क्षेत्र देखें।
It is ((x-c)2-25), so \(x-c=\pm5\) and the zeroes are (c-5), (c+5). Tip: use difference of squares.
Step 2
Why this answer is correct
The correct answer is A. (c-5) और (c+5) / (c-5) and (c+5). It is ((x-c)2-25), so \(x-c=\pm5\) and the zeroes are (c-5), (c+5). Tip: use difference of squares.
Step 3
Exam Tip
यह ((x-c)2-25) है, इसलिए \(x-c=\pm5\) और शून्यक (c-5), (c+5) हैं। टिप: वर्गों के अंतर का उपयोग करें।
In this interval the signs are (+), (-), (-), so the product is positive. Tip: the product of two negative factors is positive.
Step 2
Why this answer is correct
The correct answer is A. (x)-अक्ष के ऊपर / Above the (x)-axis. In this interval the signs are (+), (-), (-), so the product is positive. Tip: the product of two negative factors is positive.
Step 3
Exam Tip
इस अंतराल में चिह्न (+), (-), (-) हैं, इसलिए गुणनफल धनात्मक है। टिप: दो ऋणात्मक कारकों का गुणन धनात्मक होता है।
