The average of the two zeroes is (-2), so the other zero is (-9). Tip: connect the axis of symmetry with the midpoint of zeroes.
Step 2
Why this answer is correct
The correct answer is A. (-9). The average of the two zeroes is (-2), so the other zero is (-9). Tip: connect the axis of symmetry with the midpoint of zeroes.
Step 3
Exam Tip
दो शून्यकों का औसत (-2) है, इसलिए दूसरा शून्यक (-9) होगा। टिप: सममिति अक्ष को शून्यकों के मध्य से जोड़ें।
A. (x=-5) पर काटेगा और (x=2) पर स्पर्श करेगा/It crosses at (x=-5) and touches at (x=2)
Step 1
Concept
An odd-power zero gives crossing and an even-power zero gives touching. Tip: identify graph behavior from the power of the factor.
Step 2
Why this answer is correct
The correct answer is A. (x=-5) पर काटेगा और (x=2) पर स्पर्श करेगा / It crosses at (x=-5) and touches at (x=2). An odd-power zero gives crossing and an even-power zero gives touching. Tip: identify graph behavior from the power of the factor.
Step 3
Exam Tip
विषम घात वाला शून्यक कटान और सम घात वाला शून्यक स्पर्श देता है। टिप: कारक की घात से ग्राफ का व्यवहार पहचानें।
(0) lies between the two zeroes and an upward parabola stays below there. Tip: check the sign region between zeroes.
Step 2
Why this answer is correct
The correct answer is B. (x)-अक्ष के नीचे / Below the (x)-axis. (0) lies between the two zeroes and an upward parabola stays below there. Tip: check the sign region between zeroes.
Step 3
Exam Tip
(0) दोनों शून्यकों के बीच है और ऊपर खुलने वाला परवलय बीच में नीचे रहता है। टिप: शून्यकों के बीच संकेत क्षेत्र देखें।
In this interval the factor signs are (-), (+), (+), and the outside negative makes the value positive. Tip: apply the outside sign at the end.
Step 2
Why this answer is correct
The correct answer is A. ऊपर / Above. In this interval the factor signs are (-), (+), (+), and the outside negative makes the value positive. Tip: apply the outside sign at the end.
Step 3
Exam Tip
इस अंतराल में कारकों के चिह्न (-), (+), (+) हैं और बाहर का ऋण चिन्ह मान को धनात्मक बनाता है। टिप: बाहरी चिन्ह को अंत में लगाएं।
The polynomial is ((x-m)(x-(m-1))), so the zeroes are (m) and (m-1). Tip: write zeroes as ((x,0)).
Step 2
Why this answer is correct
The correct answer is A. ((m,0)) और ((m-1,0)) / ((m,0)) and ((m-1,0)). The polynomial is ((x-m)(x-(m-1))), so the zeroes are (m) and (m-1). Tip: write zeroes as ((x,0)).
Step 3
Exam Tip
बहुपद ((x-m)(x-(m-1))) है, इसलिए शून्यक (m) और (m-1) हैं। टिप: शून्यकों को ((x,0)) में लिखें।
The other zero is (4), and the average is \(\frac{-8+4}{2}=-2\). Tip: the axis of symmetry is the average of two zeroes.
Step 2
Why this answer is correct
The correct answer is A. (x=-2). The other zero is (4), and the average is \(\frac{-8+4}{2}=-2\). Tip: the axis of symmetry is the average of two zeroes.
Step 3
Exam Tip
दूसरा शून्यक (4) है और औसत \(\frac{-8+4}{2}=-2\) है। टिप: सममिति अक्ष दो शून्यकों का औसत है।
There are two distinct zeroes (1) and (-4), and both have even powers. Tip: at an even-power zero the graph usually touches.
Step 2
Why this answer is correct
The correct answer is A. दो, दोनों पर स्पर्श / Two, touches at both. There are two distinct zeroes (1) and (-4), and both have even powers. Tip: at an even-power zero the graph usually touches.
Step 3
Exam Tip
दो अलग शून्यक (1) और (-4) हैं तथा दोनों की घात सम है। टिप: सम घात वाले शून्यक पर ग्राफ सामान्यतः स्पर्श करता है।
From (x+a=0) we get (-a), and from (x-b=0) we get (b). Tip: do not count repetition among distinct zeroes.
Step 2
Why this answer is correct
The correct answer is A. (-a) और (b) / (-a) and (b). From (x+a=0) we get (-a), and from (x-b=0) we get (b). Tip: do not count repetition among distinct zeroes.
Step 3
Exam Tip
(x+a=0) से (-a) और (x-b=0) से (b) मिलता है। टिप: अलग शून्यकों में दोहराव न गिनें।
(x-2-9x-52=(x-13)(x+4)), so the zeroes are (13) and (-4). Tip: write intersection points from factors.
Step 2
Why this answer is correct
The correct answer is A. ((13,0)) और ((-4,0)) / ((13,0)) and ((-4,0)). (x-2-9x-52=(x-13)(x+4)), so the zeroes are (13) and (-4). Tip: write intersection points from factors.
Step 3
Exam Tip
(x-2-9x-52=(x-13)(x+4)) है, इसलिए शून्यक (13) और (-4) हैं। टिप: गुणनखंडों से कटान बिंदु लिखें।
The origin is also on the (x)-axis, and (x=6) 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 (6) / (0) and (6). The origin is also on the (x)-axis, and (x=6) 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=6) भी (x)-अक्ष कटान है। टिप: ((0,0)) को शून्यक (0) के रूप में गिनें।
A. क्योंकि यह ((x-5)2+4) है/Because it is ((x-5)2+4)
Step 1
Concept
((x-5)2+4) is always positive, so (p(x)=0) will not occur. Tip: adding a positive number to a square gives no real intersection.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि यह ((x-5)2+4) है / Because it is ((x-5)2+4). ((x-5)2+4) is always positive, so (p(x)=0) will not occur. Tip: adding a positive number to a square gives no real intersection.
Step 3
Exam Tip
((x-5)2+4) हमेशा धनात्मक है, इसलिए (p(x)=0) नहीं होगा। टिप: वर्ग में धनात्मक संख्या जुड़ने पर वास्तविक कटान नहीं मिलता।
B. ग्राफ (x)-अक्ष को स्पर्श करेगा/The graph will touch the (x)-axis
Step 1
Concept
((x-7)4) is an even-power factor, so the graph touches at (x=7). Tip: at an even power the graph usually turns back.
Step 2
Why this answer is correct
The correct answer is B. ग्राफ (x)-अक्ष को स्पर्श करेगा / The graph will touch the (x)-axis. ((x-7)4) is an even-power factor, so the graph touches at (x=7). Tip: at an even power the graph usually turns back.
Step 3
Exam Tip
((x-7)4) सम घात का कारक है, इसलिए (x=7) पर स्पर्श होगा। टिप: सम घात में ग्राफ आमतौर पर लौटता है।
The axis of symmetry is at the average of the zeroes, (\frac{(t-9)+(t+5)}{2}=t-2). Tip: take the midpoint even with symbols.
Step 2
Why this answer is correct
The correct answer is A. (x=t-2). The axis of symmetry is at the average of the zeroes, (\frac{(t-9)+(t+5)}{2}=t-2). Tip: take the midpoint even with symbols.
Step 3
Exam Tip
सममिति अक्ष शून्यकों के औसत पर है, (\frac{(t-9)+(t+5)}{2}=t-2)। टिप: प्रतीकों में भी मध्य मान लें।
In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check the sign of each factor separately.
Step 2
Why this answer is correct
The correct answer is B. ऋणात्मक / Negative. In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check the sign of each factor separately.
Step 3
Exam Tip
इस अंतराल में पहले दो कारक धनात्मक और तीसरा ऋणात्मक है, इसलिए गुणनफल ऋणात्मक है। टिप: प्रत्येक कारक का चिह्न अलग जांचें।
A. दूसरा (7), कटान ((4,0)), ((7,0))/Other (7), intersections ((4,0)), ((7,0))
Step 1
Concept
In the quadratic, the sum of zeroes is (11), so the other zero is (7). Tip: convert a zero into ((x,0)).
Step 2
Why this answer is correct
The correct answer is A. दूसरा (7), कटान ((4,0)), ((7,0)) / Other (7), intersections ((4,0)), ((7,0)). In the quadratic, the sum of zeroes is (11), so the other zero is (7). Tip: convert a zero into ((x,0)).
Step 3
Exam Tip
द्विघात में शून्यकों का योग (11) है, इसलिए दूसरा शून्यक (7) है। टिप: शून्यक को ((x,0)) में बदलें।
The zeroes are (-10) and (10), so the product is (-100) and the sum is (0). Tip: opposite zeroes have sum (0).
Step 2
Why this answer is correct
The correct answer is A. गुणनफल (-100), योग (0) / Product (-100), sum (0). The zeroes are (-10) and (10), so the product is (-100) and the sum is (0). Tip: opposite zeroes have sum (0).
Step 3
Exam Tip
शून्यक (-10) और (10) हैं, इसलिए गुणनफल (-100) और योग (0) है। टिप: विपरीत शून्यकों का योग (0) होता है।
A. (\left\(\frac{6}{5},0\right\)) और (\left\(-\frac{6}{5},0\right\))/(\left\(\frac{6}{5},0\right\)) and (\left\(-\frac{6}{5},0\right\))
Step 1
Concept
From \(25x^2-36=0\), \(x=\pm\frac{6}{5}\). Tip: treat \(25x^2\) as ((5x)2).
Step 2
Why this answer is correct
The correct answer is A. (\left\(\frac{6}{5},0\right\)) और (\left\(-\frac{6}{5},0\right\)) / (\left\(\frac{6}{5},0\right\)) and (\left\(-\frac{6}{5},0\right\)). From \(25x^2-36=0\), \(x=\pm\frac{6}{5}\). Tip: treat \(25x^2\) as ((5x)2).
Step 3
Exam Tip
\(25x^2-36=0\) से \(x=\pm\frac{6}{5}\) मिलता है। टिप: \(25x^2\) को ((5x)2) समझें।
Repeated points give the same (x)-values, so the distinct zeroes are (-7) and (2). Tip: count the same (x)-value once.
Step 2
Why this answer is correct
The correct answer is B. दो / Two. Repeated points give the same (x)-values, so the distinct zeroes are (-7) and (2). Tip: count the same (x)-value once.
Step 3
Exam Tip
दोहराए बिंदु समान (x)-मान देते हैं, इसलिए अलग शून्यक (-7) और (2) हैं। टिप: समान (x)-मान को एक बार गिनें।
Real zeroes are counted from (x)-axis intersections, not from the (y)-axis intercept. Tip: ((0,-20)) does not show a zero.
Step 2
Why this answer is correct
The correct answer is B. दो / Two. Real zeroes are counted from (x)-axis intersections, not from the (y)-axis intercept. Tip: ((0,-20)) does not show a zero.
Step 3
Exam Tip
वास्तविक शून्यक (x)-अक्ष कटानों से गिने जाते हैं, (y)-अक्ष कटान से नहीं। टिप: ((0,-20)) शून्यक नहीं बताता।
(x-4-625=\(x^2-25\)\(x^2+25\)), and the real zeroes are only \(\pm5\). Tip: \(x^2+25\) gives no real zero.
Step 2
Why this answer is correct
The correct answer is A. ((-5,0)) और ((5,0)) / ((-5,0)) and ((5,0)). (x-4-625=\(x^2-25\)\(x^2+25\)), and the real zeroes are only \(\pm5\). Tip: \(x^2+25\) gives no real zero.
Step 3
Exam Tip
(x-4-625=\(x^2-25\)\(x^2+25\)) है और वास्तविक शून्यक केवल \(\pm5\) हैं। टिप: \(x^2+25\) वास्तविक शून्यक नहीं देता।
For seven distinct real zeroes, the degree must be at least (7). Tip: the number of distinct zeroes cannot exceed the degree.
Step 2
Why this answer is correct
The correct answer is C. (7). For seven distinct real zeroes, the degree must be at least (7). Tip: the number of distinct zeroes cannot exceed the degree.
Step 3
Exam Tip
सात अलग वास्तविक शून्यकों के लिए घात कम से कम (7) होनी चाहिए। टिप: अलग शून्यकों की संख्या घात से अधिक नहीं हो सकती।
C. ग्राफ (x)-अक्ष को नहीं काटता/The graph does not cut the (x)-axis
Step 1
Concept
(x-2+6x+18=(x+3)2+9), so there is no real zero. Tip: an always positive form gives no intersection.
Step 2
Why this answer is correct
The correct answer is C. ग्राफ (x)-अक्ष को नहीं काटता / The graph does not cut the (x)-axis. (x-2+6x+18=(x+3)2+9), so there is no real zero. Tip: an always positive form gives no intersection.
Step 3
Exam Tip
(x-2+6x+18=(x+3)2+9) है, इसलिए वास्तविक शून्यक नहीं है। टिप: हमेशा धनात्मक रूप कटान नहीं देता।
A. यह दोनों शून्यकों का मध्य है/It is the midpoint of the two zeroes
Step 1
Concept
\(14=\frac{8+20}{2}\), 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. \(14=\frac{8+20}{2}\), so it is the midpoint of the two zeroes. Tip: the midpoint value need not be a zero.
Step 3
Exam Tip
\(14=\frac{8+20}{2}\), इसलिए यह दोनों शून्यकों का मध्य है। टिप: मध्य मान का शून्यक होना जरूरी नहीं।
(5x-2-50x+125=5(x-5)2), so it touches at (x=5). Tip: the outside (5) does not change the zero.
Step 2
Why this answer is correct
The correct answer is A. (x=5) पर स्पर्श करेगा / It will touch at (x=5). (5x-2-50x+125=5(x-5)2), so it touches at (x=5). Tip: the outside (5) does not change the zero.
Step 3
Exam Tip
(5x-2-50x+125=5(x-5)2) है, इसलिए (x=5) पर स्पर्श होगा। टिप: बाहरी (5) शून्यक नहीं बदलता।
For (x<-11), both factors are negative and the outside negative makes the value negative. Tip: check factor signs first.
Step 2
Why this answer is correct
The correct answer is B. नीचे / Below. For (x<-11), both factors are negative and the outside negative makes the value negative. Tip: check factor signs first.
Step 3
Exam Tip
(x<-11) पर दोनों कारक ऋणात्मक हैं और बाहर का ऋण चिन्ह मान को ऋणात्मक बनाता है। टिप: पहले कारकों का चिह्न देखें।
For a downward-opening parabola, values between the two zeroes are positive. Tip: (x=4) lies between the zeroes.
Step 2
Why this answer is correct
The correct answer is A. (x)-अक्ष के ऊपर / Above the (x)-axis. For a downward-opening parabola, values between the two zeroes are positive. Tip: (x=4) lies between the zeroes.
Step 3
Exam Tip
नीचे खुलने वाले परवलय में दो शून्यकों के बीच मान धनात्मक होते हैं। टिप: (x=4) दोनों शून्यकों के बीच है।
A. (18) को (8) करना होगा/(18) must be changed to (8)
Step 1
Concept
For equal distance from the (y)-axis, zeroes should be opposites, so (8) is needed with (-8). Tip: symmetric zeroes are (a) and (-a).
Step 2
Why this answer is correct
The correct answer is A. (18) को (8) करना होगा / (18) must be changed to (8). For equal distance from the (y)-axis, zeroes should be opposites, so (8) is needed with (-8). Tip: symmetric zeroes are (a) and (-a).
Step 3
Exam Tip
(y)-अक्ष से समान दूरी के लिए शून्यक विपरीत होने चाहिए, इसलिए (-8) के साथ (8) चाहिए। टिप: सममित शून्यक (a) और (-a) होते हैं।
A. दो बिंदु, (x=-4) पर स्पर्श/Two points, touching at (x=-4)
Step 1
Concept
The zeroes are (-4) and (12), and ((x+4)2) causes touching at (-4). Tip: the outside (9) does not change the zeroes.
Step 2
Why this answer is correct
The correct answer is A. दो बिंदु, (x=-4) पर स्पर्श / Two points, touching at (x=-4). The zeroes are (-4) and (12), and ((x+4)2) causes touching at (-4). Tip: the outside (9) does not change the zeroes.
Step 3
Exam Tip
शून्यक (-4) और (12) हैं तथा ((x+4)2) के कारण (-4) पर स्पर्श है। टिप: बाहरी (9) शून्यक नहीं बदलता।
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. क्योंकि इसका विविक्तकर ऋणात्मक है/Because its discriminant is negative
Step 1
Concept
The discriminant is \(f^2-4f^2=-3f^2<0\), so there are no real zeroes. Tip: a negative discriminant means no (x)-axis intersection.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि इसका विविक्तकर ऋणात्मक है / Because its discriminant is negative. The discriminant is \(f^2-4f^2=-3f^2<0\), so there are no real zeroes. Tip: a negative discriminant means no (x)-axis intersection.
Step 3
Exam Tip
विविक्तकर \(f^2-4f^2=-3f^2<0\) है, इसलिए वास्तविक शून्यक नहीं हैं। टिप: ऋणात्मक विविक्तकर का अर्थ (x)-अक्ष कटान नहीं है।
The vertex lies on the (x)-axis, so the parabola touches at ((12,0)). Tip: if the vertex has (y=0), there is one distinct zero.
Step 2
Why this answer is correct
The correct answer is B. एक / One. The vertex lies on the (x)-axis, so the parabola touches at ((12,0)). Tip: if the vertex has (y=0), there is one distinct zero.
Step 3
Exam Tip
शीर्ष (x)-अक्ष पर है, इसलिए परवलय ((12,0)) पर स्पर्श करेगा। टिप: शीर्ष का (y)-मान (0) हो तो एक अलग शून्यक होता है।
The discriminant is (256-336=-80), so there are no real zeroes. Tip: with negative discriminant a parabola does not meet the (x)-axis.
Step 2
Why this answer is correct
The correct answer is C. नहीं काटेगा / It will not cut. The discriminant is (256-336=-80), so there are no real zeroes. Tip: with negative discriminant a parabola does not meet the (x)-axis.
Step 3
Exam Tip
विविक्तकर (256-336=-80) है, इसलिए वास्तविक शून्यक नहीं हैं। टिप: ऋणात्मक विविक्तकर पर परवलय (x)-अक्ष से नहीं मिलता।
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।
The average of the two zeroes is (5), so the other zero is (11). Tip: the axis of symmetry passes through the midpoint of zeroes.
Step 2
Why this answer is correct
The correct answer is A. (11). The average of the two zeroes is (5), so the other zero is (11). Tip: the axis of symmetry passes through the midpoint of zeroes.
Step 3
Exam Tip
दो शून्यकों का औसत (5) है इसलिए दूसरा शून्यक (11) होगा। टिप: सममिति अक्ष शून्यकों के मध्य से गुजरता है।
A. (x=4) पर स्पर्श और (x=-3) पर कटान/Touches at (x=4) and crosses at (x=-3)
Step 1
Concept
An even-power zero gives touching and an odd-power zero gives crossing. Tip: identify graph behavior from the power of the factor.
Step 2
Why this answer is correct
The correct answer is A. (x=4) पर स्पर्श और (x=-3) पर कटान / Touches at (x=4) and crosses at (x=-3). An even-power zero gives touching and an odd-power zero gives crossing. Tip: identify graph behavior from the power of the factor.
Step 3
Exam Tip
सम घात वाला शून्यक स्पर्श और विषम घात वाला शून्यक कटान देता है। टिप: कारक की घात से ग्राफ का व्यवहार पहचानें।
(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।
In this interval the factor signs are (+), (+), (-), and the outside negative makes the value positive. Tip: apply the outside sign at the end.
Step 2
Why this answer is correct
The correct answer is A. ऊपर / Above. In this interval the factor signs are (+), (+), (-), and the outside negative makes the value positive. Tip: apply the outside sign at the end.
Step 3
Exam Tip
इस अंतराल में कारकों के चिह्न (+), (+), (-) हैं और बाहर का ऋण चिन्ह मान को धनात्मक बनाता है। टिप: बाहरी चिन्ह को अंत में लगाएं।
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।
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।
There are two distinct zeroes (-2) and (5), and both have even powers. Tip: at an even-power zero the graph usually touches.
Step 2
Why this answer is correct
The correct answer is A. दो, दोनों पर स्पर्श / Two, touches at both. There are two distinct zeroes (-2) and (5), and both have even powers. Tip: at an even-power zero the graph usually touches.
Step 3
Exam Tip
दो अलग शून्यक (-2) और (5) हैं तथा दोनों की घात सम है। टिप: सम घात वाले शून्यक पर ग्राफ सामान्यतः स्पर्श करता है।
From (x-c=0) we get (c), and from (x+d=0) we get (-d). Tip: do not count repetition among distinct zeroes.
Step 2
Why this answer is correct
The correct answer is A. (c) और (-d) / (c) and (-d). From (x-c=0) we get (c), and from (x+d=0) we get (-d). Tip: do not count repetition among distinct zeroes.
Step 3
Exam Tip
(x-c=0) से (c) और (x+d=0) से (-d) मिलता है। टिप: अलग शून्यकों में दोहराव न गिनें।
(x-2-13x-68=(x-17)(x+4)), so the zeroes are (17) and (-4). Tip: write intersection points from factors.
Step 2
Why this answer is correct
The correct answer is A. ((17,0)) और ((-4,0)) / ((17,0)) and ((-4,0)). (x-2-13x-68=(x-17)(x+4)), so the zeroes are (17) and (-4). Tip: write intersection points from factors.
Step 3
Exam Tip
(x-2-13x-68=(x-17)(x+4)) है, इसलिए शून्यक (17) और (-4) हैं। टिप: गुणनखंडों से कटान बिंदु लिखें।
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।
(x-2+20x+100=(x+10)2), so the touching point is ((-10,0)). Tip: change the sign in a perfect square to get the zero.
Step 2
Why this answer is correct
The correct answer is A. ((-10,0)). (x-2+20x+100=(x+10)2), so the touching point is ((-10,0)). Tip: change the sign in a perfect square to get the zero.
Step 3
Exam Tip
(x-2+20x+100=(x+10)2) है, इसलिए स्पर्श बिंदु ((-10,0)) है। टिप: पूर्ण वर्ग में चिह्न बदलकर शून्यक मिलता है।
A. क्योंकि यह ((x-7)2+4) है/Because it is ((x-7)2+4)
Step 1
Concept
((x-7)2+4) is always positive, so (p(x)=0) will not occur. Tip: adding a positive number to a square gives no real intersection.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि यह ((x-7)2+4) है / Because it is ((x-7)2+4). ((x-7)2+4) is always positive, so (p(x)=0) will not occur. Tip: adding a positive number to a square gives no real intersection.
Step 3
Exam Tip
((x-7)2+4) हमेशा धनात्मक है इसलिए (p(x)=0) नहीं होगा। टिप: वर्ग में धनात्मक संख्या जुड़ने पर वास्तविक कटान नहीं मिलता।
A. ग्राफ (x)-अक्ष को काटेगा/The graph will cross the (x)-axis
Step 1
Concept
((x-3)5) is an odd-power factor, so the graph crosses at (x=3). Tip: at an odd power the graph usually crosses the axis.
Step 2
Why this answer is correct
The correct answer is A. ग्राफ (x)-अक्ष को काटेगा / The graph will cross the (x)-axis. ((x-3)5) is an odd-power factor, so the graph crosses at (x=3). Tip: at an odd power the graph usually crosses the axis.
Step 3
Exam Tip
((x-3)5) विषम घात का कारक है, इसलिए (x=3) पर कटान होगा। टिप: विषम घात में ग्राफ सामान्यतः अक्ष पार करता है।
The axis of symmetry is at the average of the zeroes, (\frac{(q-11)+(q+7)}{2}=q-2). Tip: take the midpoint even with symbols.
Step 2
Why this answer is correct
The correct answer is A. (x=q-2). The axis of symmetry is at the average of the zeroes, (\frac{(q-11)+(q+7)}{2}=q-2). Tip: take the midpoint even with symbols.
Step 3
Exam Tip
सममिति अक्ष शून्यकों के औसत पर है, (\frac{(q-11)+(q+7)}{2}=q-2)। टिप: प्रतीकों में भी मध्य मान लें।
In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check the sign of each factor separately.
Step 2
Why this answer is correct
The correct answer is B. ऋणात्मक / Negative. In this interval the first two factors are positive and the third is negative, so the product is negative. Tip: check the sign of each factor separately.
Step 3
Exam Tip
इस अंतराल में पहले दो कारक धनात्मक और तीसरा ऋणात्मक है, इसलिए गुणनफल ऋणात्मक है। टिप: प्रत्येक कारक का चिह्न अलग जांचें।
A. दूसरा (7), कटान ((6,0)), ((7,0))/Other (7), intersections ((6,0)), ((7,0))
Step 1
Concept
In the quadratic, the sum of zeroes is (13), so the other zero is (7). Tip: convert a zero into ((x,0)).
Step 2
Why this answer is correct
The correct answer is A. दूसरा (7), कटान ((6,0)), ((7,0)) / Other (7), intersections ((6,0)), ((7,0)). In the quadratic, the sum of zeroes is (13), so the other zero is (7). Tip: convert a zero into ((x,0)).
Step 3
Exam Tip
द्विघात में शून्यकों का योग (13) है, इसलिए दूसरा शून्यक (7) है। टिप: शून्यक को ((x,0)) में बदलें।
The zeroes are (-12) and (12), so the product is (-144) and the sum is (0). Tip: opposite zeroes have sum (0).
Step 2
Why this answer is correct
The correct answer is A. गुणनफल (-144), योग (0) / Product (-144), sum (0). The zeroes are (-12) and (12), so the product is (-144) and the sum is (0). Tip: opposite zeroes have sum (0).
Step 3
Exam Tip
शून्यक (-12) और (12) हैं, इसलिए गुणनफल (-144) और योग (0) है। टिप: विपरीत शून्यकों का योग (0) होता है।
A. (\left\(\frac{7}{6},0\right\)) और (\left\(-\frac{7}{6},0\right\))/(\left\(\frac{7}{6},0\right\)) and (\left\(-\frac{7}{6},0\right\))
Step 1
Concept
From \(36x^2-49=0\), \(x=\pm\frac{7}{6}\). Tip: treat \(36x^2\) as ((6x)2).
Step 2
Why this answer is correct
The correct answer is A. (\left\(\frac{7}{6},0\right\)) और (\left\(-\frac{7}{6},0\right\)) / (\left\(\frac{7}{6},0\right\)) and (\left\(-\frac{7}{6},0\right\)). From \(36x^2-49=0\), \(x=\pm\frac{7}{6}\). Tip: treat \(36x^2\) as ((6x)2).
Step 3
Exam Tip
\(36x^2-49=0\) से \(x=\pm\frac{7}{6}\) मिलता है। टिप: \(36x^2\) को ((6x)2) समझें।
Repeated points give the same (x)-values, so the distinct zeroes are (-11) and (4). Tip: count the same (x)-value once.
Step 2
Why this answer is correct
The correct answer is B. दो / Two. Repeated points give the same (x)-values, so the distinct zeroes are (-11) and (4). Tip: count the same (x)-value once.
Step 3
Exam Tip
दोहराए बिंदु समान (x)-मान देते हैं, इसलिए अलग शून्यक (-11) और (4) हैं। टिप: समान (x)-मान को एक बार गिनें।
Real zeroes are counted from (x)-axis intersections, not from the (y)-axis intercept. Tip: ((0,25)) does not show a zero.
Step 2
Why this answer is correct
The correct answer is B. दो / Two. Real zeroes are counted from (x)-axis intersections, not from the (y)-axis intercept. Tip: ((0,25)) does not show a zero.
Step 3
Exam Tip
वास्तविक शून्यक (x)-अक्ष कटानों से गिने जाते हैं, (y)-अक्ष कटान से नहीं। टिप: ((0,25)) शून्यक नहीं बताता।
(x-4-1296=\(x^2-36\)\(x^2+36\)), and the real zeroes are only \(\pm6\). Tip: \(x^2+36\) gives no real zero.
Step 2
Why this answer is correct
The correct answer is A. ((-6,0)) और ((6,0)) / ((-6,0)) and ((6,0)). (x-4-1296=\(x^2-36\)\(x^2+36\)), and the real zeroes are only \(\pm6\). Tip: \(x^2+36\) gives no real zero.
Step 3
Exam Tip
(x-4-1296=\(x^2-36\)\(x^2+36\)) है और वास्तविक शून्यक केवल \(\pm6\) हैं। टिप: \(x^2+36\) वास्तविक शून्यक नहीं देता।
For eight distinct real zeroes, the degree must be at least (8). Tip: the number of distinct zeroes cannot exceed the degree.
Step 2
Why this answer is correct
The correct answer is C. (8). For eight distinct real zeroes, the degree must be at least (8). Tip: the number of distinct zeroes cannot exceed the degree.
Step 3
Exam Tip
आठ अलग वास्तविक शून्यकों के लिए घात कम से कम (8) होनी चाहिए। टिप: अलग शून्यकों की संख्या घात से अधिक नहीं हो सकती।
C. ग्राफ (x)-अक्ष को नहीं काटता/The graph does not cut the (x)-axis
Step 1
Concept
(x-2+16x+80=(x+8)2+16), so there is no real zero. Tip: an always positive form gives no intersection.
Step 2
Why this answer is correct
The correct answer is C. ग्राफ (x)-अक्ष को नहीं काटता / The graph does not cut the (x)-axis. (x-2+16x+80=(x+8)2+16), so there is no real zero. Tip: an always positive form gives no intersection.
Step 3
Exam Tip
(x-2+16x+80=(x+8)2+16) है, इसलिए वास्तविक शून्यक नहीं है। टिप: हमेशा धनात्मक रूप कटान नहीं देता।
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।
(6x-2-72x+216=6(x-6)2), so it touches at (x=6). Tip: the outside (6) does not change the zero.
Step 2
Why this answer is correct
The correct answer is A. (x=6) पर स्पर्श करेगा / It will touch at (x=6). (6x-2-72x+216=6(x-6)2), so it touches at (x=6). Tip: the outside (6) does not change the zero.
Step 3
Exam Tip
(6x-2-72x+216=6(x-6)2) है, इसलिए (x=6) पर स्पर्श होगा। टिप: बाहरी (6) शून्यक नहीं बदलता।
For (x<-13), both factors are negative and the outside negative makes the value negative. Tip: check factor signs first.
Step 2
Why this answer is correct
The correct answer is B. नीचे / Below. For (x<-13), both factors are negative and the outside negative makes the value negative. Tip: check factor signs first.
Step 3
Exam Tip
(x<-13) पर दोनों कारक ऋणात्मक हैं और बाहर का ऋण चिन्ह मान को ऋणात्मक बनाता है। टिप: पहले कारकों का चिह्न देखें।
For a downward-opening parabola, values between the two zeroes are positive. Tip: (x=2) lies between the zeroes.
Step 2
Why this answer is correct
The correct answer is A. (x)-अक्ष के ऊपर / Above the (x)-axis. For a downward-opening parabola, values between the two zeroes are positive. Tip: (x=2) lies between the zeroes.
Step 3
Exam Tip
नीचे खुलने वाले परवलय में दो शून्यकों के बीच मान धनात्मक होते हैं। टिप: (x=2) दोनों शून्यकों के बीच 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. दो बिंदु, (x=-5) पर स्पर्श/Two points, touching at (x=-5)
Step 1
Concept
The zeroes are (-5) and (14), and ((x+5)2) causes touching at (-5). Tip: the outside (11) does not change the zeroes.
Step 2
Why this answer is correct
The correct answer is A. दो बिंदु, (x=-5) पर स्पर्श / Two points, touching at (x=-5). The zeroes are (-5) and (14), and ((x+5)2) causes touching at (-5). Tip: the outside (11) does not change the zeroes.
Step 3
Exam Tip
शून्यक (-5) और (14) हैं तथा ((x+5)2) के कारण (-5) पर स्पर्श है। टिप: बाहरी (11) शून्यक नहीं बदलता।
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. क्योंकि इसका विविक्तकर ऋणात्मक है/Because its discriminant is negative
Step 1
Concept
The discriminant is \(g^2-4g^2=-3g^2<0\), so there are no real zeroes. Tip: a negative discriminant means no (x)-axis intersection.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि इसका विविक्तकर ऋणात्मक है / Because its discriminant is negative. The discriminant is \(g^2-4g^2=-3g^2<0\), so there are no real zeroes. Tip: a negative discriminant means no (x)-axis intersection.
Step 3
Exam Tip
विविक्तकर \(g^2-4g^2=-3g^2<0\) है, इसलिए वास्तविक शून्यक नहीं हैं। टिप: ऋणात्मक विविक्तकर का अर्थ (x)-अक्ष कटान नहीं है।
The vertex lies on the (x)-axis, so the parabola touches at ((-14,0)). Tip: if the vertex has (y=0), there is one distinct zero.
Step 2
Why this answer is correct
The correct answer is B. एक / One. The vertex lies on the (x)-axis, so the parabola touches at ((-14,0)). Tip: if the vertex has (y=0), there is one distinct zero.
Step 3
Exam Tip
शीर्ष (x)-अक्ष पर है, इसलिए परवलय ((-14,0)) पर स्पर्श करेगा। टिप: शीर्ष का (y)-मान (0) हो तो एक अलग शून्यक होता है।
The discriminant is (400-460=-60), so there are no real zeroes. Tip: with negative discriminant a parabola does not meet the (x)-axis.
Step 2
Why this answer is correct
The correct answer is C. नहीं काटेगा / It will not cut. The discriminant is (400-460=-60), so there are no real zeroes. Tip: with negative discriminant a parabola does not meet the (x)-axis.
Step 3
Exam Tip
विविक्तकर (400-460=-60) है, इसलिए वास्तविक शून्यक नहीं हैं। टिप: ऋणात्मक विविक्तकर पर परवलय (x)-अक्ष से नहीं मिलता।
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।
A. बराबर शून्यक (4) और (4)/Equal zeroes (4) and (4)
Step 1
Concept
When a quadratic graph touches at one point its two zeroes are equal. Treat it as a repeated zero in exams.
Step 2
Why this answer is correct
The correct answer is A. बराबर शून्यक (4) और (4) / Equal zeroes (4) and (4). When a quadratic graph touches at one point its two zeroes are equal. Treat it as a repeated zero in exams.
Step 3
Exam Tip
द्विघात ग्राफ एक बिंदु पर छूता है तो दोनों शून्यक समान होते हैं। परीक्षा में इसे दोहराया शून्यक मानें।
A. ग्राफ (x)-अक्ष को ((2,0)) पर मिलता है/The graph meets the (x)-axis at ((2,0))
Step 1
Concept
(p(2)=0) means (y=0) at (x=2). So the point lies on the (x)-axis.
Step 2
Why this answer is correct
The correct answer is A. ग्राफ (x)-अक्ष को ((2,0)) पर मिलता है / The graph meets the (x)-axis at ((2,0)). (p(2)=0) means (y=0) at (x=2). So the point lies on the (x)-axis.
Step 3
Exam Tip
(p(2)=0) का अर्थ (x=2) पर (y=0) है। इसलिए बिंदु (x)-अक्ष पर होगा।
A. यह (x)-अक्ष को नहीं काटता/It does not cut the (x)-axis
Step 1
Concept
\(x^2+4\) is always positive so (y=0) never occurs. Count a zero only when the graph meets the axis.
Step 2
Why this answer is correct
The correct answer is A. यह (x)-अक्ष को नहीं काटता / It does not cut the (x)-axis. \(x^2+4\) is always positive so (y=0) never occurs. Count a zero only when the graph meets the axis.
Step 3
Exam Tip
\(x^2+4\) हमेशा धनात्मक है इसलिए (y=0) नहीं होता। ग्राफ से शून्यक तभी मानें जब अक्ष से मिलन हो।
A. (0) बहुपद का शून्यक है/(0) is a zero of the polynomial
Step 1
Concept
Meeting at the origin shows (p(0)=0). It only confirms that (0) is a zero.
Step 2
Why this answer is correct
The correct answer is A. (0) बहुपद का शून्यक है / (0) is a zero of the polynomial. Meeting at the origin shows (p(0)=0). It only confirms that (0) is a zero.
Step 3
Exam Tip
मूल बिंदु पर मिलना (p(0)=0) बताता है। इससे केवल (0) का शून्यक होना निश्चित है।
A. दो भिन्न वास्तविक शून्यक/Two distinct real zeroes
Step 1
Concept
Two separate intersections give two distinct real zeroes. Different (x)-intercepts mean different zeroes.
Step 2
Why this answer is correct
The correct answer is A. दो भिन्न वास्तविक शून्यक / Two distinct real zeroes. Two separate intersections give two distinct real zeroes. Different (x)-intercepts mean different zeroes.
Step 3
Exam Tip
दो अलग कटान दो अलग वास्तविक शून्यक देते हैं। ग्राफ में अलग (x)-प्रतिच्छेद अलग शून्यक होते हैं।
The repeated factor gives equal zero (-2). A quadratic parabola generally touches the (x)-axis at such a point.
Step 2
Why this answer is correct
The correct answer is A. ((-2,0)) पर छुएगा / It will touch at ((-2,0)). The repeated factor gives equal zero (-2). A quadratic parabola generally touches the (x)-axis at such a point.
Step 3
Exam Tip
दोहराए गुणनखंड से समान शून्यक (-2) मिलता है। द्विघात परवलय ऐसे बिंदु पर सामान्यतः (x)-अक्ष को छूता है।
A. दोनों वास्तविक शून्यक समान हैं/Both real zeroes are equal
Step 1
Concept
When the vertex lies on the (x)-axis, the parabola touches the axis at one point. Hence the zeroes are equal.
Step 2
Why this answer is correct
The correct answer is A. दोनों वास्तविक शून्यक समान हैं / Both real zeroes are equal. When the vertex lies on the (x)-axis, the parabola touches the axis at one point. Hence the zeroes are equal.
Step 3
Exam Tip
शीर्ष (x)-अक्ष पर होने पर परवलय अक्ष को एक ही बिंदु पर छूता है। इसलिए शून्यक समान होते हैं।
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।
The origin also lies on the (x)-axis so (0) is also a zero. Write the (x)-values of all intercepts.
Step 2
Why this answer is correct
The correct answer is A. शून्यक (-2,0,5) हैं / The zeroes are (-2,0,5). The origin also lies on the (x)-axis so (0) is also a zero. Write the (x)-values of all intercepts.
Step 3
Exam Tip
मूल बिंदु भी (x)-अक्ष पर है इसलिए (0) भी शून्यक है। सभी (x)-प्रतिच्छेदों के (x)-मान लिखें।
(x-2-6x+9=(x-3)2), so (3) is a repeated zero. A repeated zero appears as touching on the graph.
Step 2
Why this answer is correct
The correct answer is A. ((3,0)) पर छुएगा / It will touch at ((3,0)). (x-2-6x+9=(x-3)2), so (3) is a repeated zero. A repeated zero appears as touching on the graph.
Step 3
Exam Tip
(x-2-6x+9=(x-3)2) है इसलिए शून्यक (3) दोहराया है। दोहराया शून्यक ग्राफ में छूने जैसा दिखता है।
A. डिग्री कम से कम (4) होगी/The degree is at least (4)
Step 1
Concept
Four distinct real zeroes need degree at least four. The number of zeroes cannot exceed the degree.
Step 2
Why this answer is correct
The correct answer is A. डिग्री कम से कम (4) होगी / The degree is at least (4). Four distinct real zeroes need degree at least four. The number of zeroes cannot exceed the degree.
Step 3
Exam Tip
चार अलग वास्तविक शून्यक के लिए डिग्री कम से कम चार चाहिए। शून्यकों की संख्या डिग्री से अधिक नहीं होती।
A. पूरा ग्राफ (x)-अक्ष है/The whole graph is the (x)-axis
Step 1
Concept
(p(x)=0) gives (y=0) for every (x). Therefore the whole (x)-axis is the graph.
Step 2
Why this answer is correct
The correct answer is A. पूरा ग्राफ (x)-अक्ष है / The whole graph is the (x)-axis. (p(x)=0) gives (y=0) for every (x). Therefore the whole (x)-axis is the graph.
Step 3
Exam Tip
(p(x)=0) हर (x) के लिए (y=0) देता है। इसलिए पूरा (x)-अक्ष ग्राफ 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. (k(x-m)(x-n)), जहाँ \(k\neq0\)/(k(x-m)(x-n)), where \(k\neq0\)
Step 1
Concept
For zeroes (m) and (n), the factors are ((x-m)) and ((x-n)). A non-zero multiplier does not change zeroes.
Step 2
Why this answer is correct
The correct answer is A. (k(x-m)(x-n)), जहाँ \(k\neq0\) / (k(x-m)(x-n)), where \(k\neq0\). For zeroes (m) and (n), the factors are ((x-m)) and ((x-n)). A non-zero multiplier does not change zeroes.
Step 3
Exam Tip
शून्यक (m) और (n) के लिए गुणनखंड ((x-m)) और ((x-n)) होते हैं। गैर शून्य गुणक शून्यक नहीं बदलता।
A. \(\frac{1}{2}\) शून्यक है/\(\frac{1}{2}\) is a zero
Step 1
Concept
A zero can be a fraction and touching is enough. The key point is (p\left\(\frac{1}{2}\right\)=0).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\) शून्यक है / \(\frac{1}{2}\) is a zero. A zero can be a fraction and touching is enough. The key point is (p\left\(\frac{1}{2}\right\)=0).
Step 3
Exam Tip
शून्यक भिन्न भी हो सकता है और छूना पर्याप्त है। जरूरी बात (p\left\(\frac{1}{2}\right\)=0) है।
A. हर (y)-प्रतिच्छेद शून्यक होता है/Every (y)-intercept is a zero
Step 1
Concept
A zero is related to the (x)-axis not a general (y)-intercept. A (y)-intercept gives a zero only if it is the origin.
Step 2
Why this answer is correct
The correct answer is A. हर (y)-प्रतिच्छेद शून्यक होता है / Every (y)-intercept is a zero. A zero is related to the (x)-axis not a general (y)-intercept. A (y)-intercept gives a zero only if it is the origin.
Step 3
Exam Tip
शून्यक (x)-अक्ष से संबंधित है न कि सामान्य (y)-प्रतिच्छेद से। (y)-प्रतिच्छेद तभी शून्यक देगा जब वह मूल बिंदु हो।
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।
The solutions of (p(x)=0) are where the graph meets the (x)-axis. So both given (x)-values are solutions.
Step 2
Why this answer is correct
The correct answer is A. (-7) और (-2) / (-7) and (-2). The solutions of (p(x)=0) are where the graph meets the (x)-axis. So both given (x)-values are solutions.
Step 3
Exam Tip
(p(x)=0) के हल वही हैं जहाँ ग्राफ (x)-अक्ष से मिलता है। इसलिए दिए गए दोनों (x)-मान हल हैं।
A. वास्तविक शून्यकों की संख्या प्रतिच्छेदों की संख्या के बराबर होती है/The number of real zeroes equals the number of intersection points
Step 1
Concept
Geometrically each (x)-axis intersection gives one real zero. A quadratic may have (0), (1), or (2) real zeroes.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक शून्यकों की संख्या प्रतिच्छेदों की संख्या के बराबर होती है / The number of real zeroes equals the number of intersection points. Geometrically each (x)-axis intersection gives one real zero. A quadratic may have (0), (1), or (2) real zeroes.
Step 3
Exam Tip
ज्यामितीय अर्थ में हर (x)-अक्ष प्रतिच्छेद एक वास्तविक शून्यक देता है। द्विघात में वास्तविक शून्यक (0), (1), या (2) हो सकते हैं।
A. वे \(a+\sqrt{7}\) और \(a-\sqrt{7}\) हैं/They are \(a+\sqrt{7}\) and \(a-\sqrt{7}\)
Step 1
Concept
(p(x)=(x-a)2-7), so \(x=a\pm\sqrt{7}\). Recognizing a perfect-square form saves time in hard questions.
Step 2
Why this answer is correct
The correct answer is A. वे \(a+\sqrt{7}\) और \(a-\sqrt{7}\) हैं / They are \(a+\sqrt{7}\) and \(a-\sqrt{7}\). (p(x)=(x-a)2-7), so \(x=a\pm\sqrt{7}\). Recognizing a perfect-square form saves time in hard questions.
Step 3
Exam Tip
(p(x)=(x-a)2-7), इसलिए \(x=a\pm\sqrt{7}\) है। पूर्ण वर्ग रूप पहचानना कठिन प्रश्नों में समय बचाता 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।
\(\alpha+\beta=10\) and \(\alpha\beta=25-6=19\), so \(\alpha^2+\beta^2=100-38=62\). Use (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta).
Step 2
Why this answer is correct
The correct answer is A. (62). \(\alpha+\beta=10\) and \(\alpha\beta=25-6=19\), so \(\alpha^2+\beta^2=100-38=62\). Use (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta).
Step 3
Exam Tip
\(\alpha+\beta=10\) और \(\alpha\beta=25-6=19\), इसलिए \(\alpha^2+\beta^2=100-38=62\)। पहचान (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta) उपयोग करें।
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।
After removing the common factor, we get \(x^2-6x+7\), and (D=36-28=8). Since (D) is positive and not a perfect square, the zeroes are real irrational.
Step 2
Why this answer is correct
The correct answer is B. वास्तविक और अपरिमेय / Real and irrational. After removing the common factor, we get \(x^2-6x+7\), and (D=36-28=8). Since (D) is positive and not a perfect square, the zeroes are real irrational.
Step 3
Exam Tip
सामान्य गुणनखंड हटाने पर \(x^2-6x+7\) मिलता है और (D=36-28=8)। (D) धनात्मक अपूर्ण वर्ग है, इसलिए शून्यक वास्तविक अपरिमेय हैं।
\(\sqrt{12}=2\sqrt{3}\), so the sum is \(3\sqrt{3}\). In a monic polynomial, the coefficient of (x) is the negative of the sum of zeroes.
Step 2
Why this answer is correct
The correct answer is A. \(-3\sqrt{3}\). \(\sqrt{12}=2\sqrt{3}\), so the sum is \(3\sqrt{3}\). In a monic polynomial, the coefficient of (x) is the negative of the sum of zeroes.
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\), इसलिए योग \(3\sqrt{3}\) है। एकक बहुपद में (x) का गुणांक शून्यकों के योग का ऋणात्मक होता है।
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. 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।
\(\alpha+\beta=4\) 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।
For \(x^2-8x+3\), (D=64-12=52), positive and not a perfect square. The other options give equal rational, non-real, or rational zeroes.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-8x+3\). For \(x^2-8x+3\), (D=64-12=52), positive and not a perfect square. The other options give equal rational, non-real, or rational zeroes.
Step 3
Exam Tip
\(x^2-8x+3\) के लिए (D=64-12=52), जो धनात्मक अपूर्ण वर्ग है। बाकी विकल्पों में शून्यक समान परिमेय, अवास्तविक या परिमेय हैं।
\(\alpha+\beta=2\) 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. (p(x)) के शून्यक परिमेय हैं और (q(x)) के शून्यक अपरिमेय वास्तविक हैं/(p(x)) has rational zeroes and (q(x)) has irrational real zeroes
Step 1
Concept
For (p(x)), (D=4+32=36), a perfect square. For (q(x)), (D=4+28=32), positive and not a perfect square.
Step 2
Why this answer is correct
The correct answer is A. (p(x)) के शून्यक परिमेय हैं और (q(x)) के शून्यक अपरिमेय वास्तविक हैं / (p(x)) has rational zeroes and (q(x)) has irrational real zeroes. For (p(x)), (D=4+32=36), a perfect square. For (q(x)), (D=4+28=32), positive and not a perfect square.
Step 3
Exam Tip
(p(x)) के लिए (D=4+32=36) पूर्ण वर्ग है। (q(x)) के लिए (D=4+28=32) धनात्मक अपूर्ण वर्ग है।
By the formula, \(x=\frac{8\pm\sqrt{64-8}}{4}=2\pm\frac{\sqrt{14}}{2}\). Divide the whole numerator by the denominator carefully.
Step 2
Why this answer is correct
The correct answer is A. \(2\pm\frac{\sqrt{14}}{2}\). By the formula, \(x=\frac{8\pm\sqrt{64-8}}{4}=2\pm\frac{\sqrt{14}}{2}\). Divide the whole numerator by the denominator carefully.
Step 3
Exam Tip
सूत्र से \(x=\frac{8\pm\sqrt{64-8}}{4}=2\pm\frac{\sqrt{14}}{2}\) है। हर से भाग देते समय पूरे अंश को बाँटें।
\(\alpha+\beta=2\) 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।
\(\sqrt{8}=2\sqrt{2}\), 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।
Using the formula, \(x=\frac{2\sqrt{3}\pm\sqrt{12+4}}{2}=\sqrt{3}\pm2\). Simplify \(\sqrt{16}=4\) carefully.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{3}\pm2\). Using the formula, \(x=\frac{2\sqrt{3}\pm\sqrt{12+4}}{2}=\sqrt{3}\pm2\). Simplify \(\sqrt{16}=4\) carefully.
Step 3
Exam Tip
सूत्र से \(x=\frac{2\sqrt{3}\pm\sqrt{12+4}}{2}=\sqrt{3}\pm2\)। \(\sqrt{16}=4\) को ध्यान से सरल करें।
The product is \(ab=\sqrt{2}\cdot\sqrt{18}=\sqrt{36}=6\). In radical multiplication, simplify the product inside the root first.
Step 2
Why this answer is correct
The correct answer is A. (6). The product is \(ab=\sqrt{2}\cdot\sqrt{18}=\sqrt{36}=6\). In radical multiplication, simplify the product inside the root first.
Step 3
Exam Tip
गुणनफल \(ab=\sqrt{2}\cdot\sqrt{18}=\sqrt{36}=6\) है। मूलों के गुणन में पहले अंदर के गुणनफल को सरल करें।
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।
(\(\alpha-\beta\)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।
A. दूसरा \(\sqrt{5}\), \(k=\sqrt{5}\)/Other \(\sqrt{5}\), \(k=\sqrt{5}\)
Step 1
Concept
The product is (5), so the other zero is \(\frac{5}{\sqrt{5}}=\sqrt{5}\). The sum is \(2\sqrt{5}=2k\), hence \(k=\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is A. दूसरा \(\sqrt{5}\), \(k=\sqrt{5}\) / Other \(\sqrt{5}\), \(k=\sqrt{5}\). The product is (5), so the other zero is \(\frac{5}{\sqrt{5}}=\sqrt{5}\). The sum is \(2\sqrt{5}=2k\), hence \(k=\sqrt{5}\).
Step 3
Exam Tip
गुणनफल (5) है, इसलिए दूसरा शून्यक \(\frac{5}{\sqrt{5}}=\sqrt{5}\) होगा। योग \(2\sqrt{5}=2k\), अतः \(k=\sqrt{5}\) है।
A. \(b^2-4c\) धनात्मक अपूर्ण वर्ग हो/\(b^2-4c\) is positive and not a perfect square
Step 1
Concept
For real zeroes, the discriminant must be positive, and for irrational zeroes it must not be a perfect square. This is the key check for quadratics with rational coefficients.
Step 2
Why this answer is correct
The correct answer is A. \(b^2-4c\) धनात्मक अपूर्ण वर्ग हो / \(b^2-4c\) is positive and not a perfect square. For real zeroes, the discriminant must be positive, and for irrational zeroes it must not be a perfect square. This is the key check for quadratics with rational coefficients.
Step 3
Exam Tip
वास्तविक शून्यकों के लिए विविक्तकर धनात्मक चाहिए और अपरिमेय शून्यकों के लिए वह पूर्ण वर्ग नहीं होना चाहिए। परिमेय गुणांकों वाले द्विघात में यही मुख्य जाँच है।
\(\alpha+\beta=2\) 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।
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।
A. दोनों परिमेय वास्तविक हैं/Both are rational real
Step 1
Concept
From \(x^2-16=0\), \(x=\pm4\), which are rational real. Not every square-root type question gives irrational roots.
Step 2
Why this answer is correct
The correct answer is A. दोनों परिमेय वास्तविक हैं / Both are rational real. From \(x^2-16=0\), \(x=\pm4\), which are rational real. Not every square-root type question gives irrational roots.
Step 3
Exam Tip
\(x^2-16=0\) से \(x=\pm4\), जो परिमेय वास्तविक हैं। हर वर्गमूल वाला प्रश्न अपरिमेय नहीं होता।
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।
A. \(-\sqrt{7}+1\) और \(-\sqrt{7}-1\)/\(-\sqrt{7}+1\) and \(-\sqrt{7}-1\)
Step 1
Concept
Using the formula, \(x=\frac{-2\sqrt{7}\pm2}{2}=-\sqrt{7}\pm1\). Simplifying the discriminant first gives a clean answer.
Step 2
Why this answer is correct
The correct answer is A. \(-\sqrt{7}+1\) और \(-\sqrt{7}-1\) / \(-\sqrt{7}+1\) and \(-\sqrt{7}-1\). Using the formula, \(x=\frac{-2\sqrt{7}\pm2}{2}=-\sqrt{7}\pm1\). Simplifying the discriminant first gives a clean answer.
Step 3
Exam Tip
सूत्र से \(x=\frac{-2\sqrt{7}\pm2}{2}=-\sqrt{7}\pm1\)। पहले विविक्तकर सरल करने से उत्तर साफ मिलता है।
A. ऐसा कोई वास्तविक (n) नहीं है/No such real (n) exists
Step 1
Concept
For equal zeroes, (D=0), so (4-4n=0) and (n=1). Then the zero is (1), which is not irrational.
Step 2
Why this answer is correct
The correct answer is A. ऐसा कोई वास्तविक (n) नहीं है / No such real (n) exists. For equal zeroes, (D=0), so (4-4n=0) and (n=1). Then the zero is (1), which is not irrational.
Step 3
Exam Tip
समान शून्यकों के लिए (D=0), यानी (4-4n=0), इसलिए (n=1)। तब शून्यक (1) है, जो अपरिमेय नहीं है।
With rational coefficients, the conjugate of the irrational part is also a zero. Hence \(\frac{3-\sqrt{5}}{2}\) is the other zero.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3-\sqrt{5}}{2}\). With rational coefficients, the conjugate of the irrational part is also a zero. Hence \(\frac{3-\sqrt{5}}{2}\) is the other zero.
Step 3
Exam Tip
परिमेय गुणांकों में अपरिमेय भाग का संयुग्मी भी शून्यक होता है। इसलिए \(\frac{3-\sqrt{5}}{2}\) दूसरा शून्यक है।
A. दो भिन्न वास्तविक अपरिमेय/Two distinct real irrational
Step 1
Concept
(D=\(2\sqrt{2}\)2-4=8-4=4), and the zeroes are \(\sqrt{2}\pm1\). They are real and irrational.
Step 2
Why this answer is correct
The correct answer is A. दो भिन्न वास्तविक अपरिमेय / Two distinct real irrational. (D=\(2\sqrt{2}\)2-4=8-4=4), and the zeroes are \(\sqrt{2}\pm1\). They are real and irrational.
Step 3
Exam Tip
(D=\(2\sqrt{2}\)2-4=8-4=4) है और शून्यक \(\sqrt{2}\pm1\) हैं। ये वास्तविक और अपरिमेय हैं।
The sum is \(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\). Simplify radicals before giving the final answer.
Step 2
Why this answer is correct
The correct answer is A. \(3\sqrt{2}\). The sum is \(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\). Simplify radicals before giving the final answer.
Step 3
Exam Tip
योग \(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\) है। मूलों को सरल करके ही अंतिम उत्तर दें।
A. (p(x)) के शून्यक परिमेय और (q(x)) के अपरिमेय वास्तविक हैं/(p(x)) has rational zeroes and (q(x)) has irrational real zeroes
Step 1
Concept
For (p(x)), (D=121-96=25), a perfect square. For (q(x)), (D=121-92=29), positive and not a perfect square.
Step 2
Why this answer is correct
The correct answer is A. (p(x)) के शून्यक परिमेय और (q(x)) के अपरिमेय वास्तविक हैं / (p(x)) has rational zeroes and (q(x)) has irrational real zeroes. For (p(x)), (D=121-96=25), a perfect square. For (q(x)), (D=121-92=29), positive and not a perfect square.
Step 3
Exam Tip
(p(x)) के लिए (D=121-96=25) पूर्ण वर्ग है। (q(x)) के लिए (D=121-92=29) धनात्मक अपूर्ण वर्ग है।
A. शून्यकों का गुणनफल \(-3\sqrt{2}\) है/The product of zeroes is \(-3\sqrt{2}\)
Step 1
Concept
In a monic quadratic, the constant term is the product of zeroes. Here \(\alpha\beta=-3\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. शून्यकों का गुणनफल \(-3\sqrt{2}\) है / The product of zeroes is \(-3\sqrt{2}\). In a monic quadratic, the constant term is the product of zeroes. Here \(\alpha\beta=-3\sqrt{2}\).
Step 3
Exam Tip
एकक द्विघात में स्थिर पद शून्यकों का गुणनफल होता है। यहाँ \(\alpha\beta=-3\sqrt{2}\) 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 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. शून्यक \(6+\sqrt{5}\) और \(6-\sqrt{5}\)/Zeroes \(6+\sqrt{5}\) and \(6-\sqrt{5}\)
Step 1
Concept
With rational coefficients, irrational parts occur in conjugate pairs. Only \(6+\sqrt{5}\) and \(6-\sqrt{5}\) have both rational sum and rational product.
Step 2
Why this answer is correct
The correct answer is A. शून्यक \(6+\sqrt{5}\) और \(6-\sqrt{5}\) / Zeroes \(6+\sqrt{5}\) and \(6-\sqrt{5}\). With rational coefficients, irrational parts occur in conjugate pairs. Only \(6+\sqrt{5}\) and \(6-\sqrt{5}\) have both rational sum and rational product.
Step 3
Exam Tip
परिमेय गुणांकों में अपरिमेय भाग संयुग्मी जोड़े में आता है। केवल \(6+\sqrt{5}\) और \(6-\sqrt{5}\) का योग और गुणनफल दोनों परिमेय हैं।
(\(1+\sqrt{3}\)2-2\(1+\sqrt{3}\)-2=1+2\sqrt{3}+3-2-2\sqrt{3}-2=0). Do not forget the middle term while expanding the square.
Step 2
Why this answer is correct
The correct answer is A. (0). (\(1+\sqrt{3}\)2-2\(1+\sqrt{3}\)-2=1+2\sqrt{3}+3-2-2\sqrt{3}-2=0). Do not forget the middle term while expanding the square.
Step 3
Exam Tip
(\(1+\sqrt{3}\)2-2\(1+\sqrt{3}\)-2=1+2\sqrt{3}+3-2-2\sqrt{3}-2=0)। वर्ग खोलते समय बीच का पद न भूलें।
Since (p\(1+\sqrt{3}\)=0), \(1+\sqrt{3}\) is a zero. To prove a number is a zero, show that the polynomial value is (0).
Step 2
Why this answer is correct
The correct answer is A. यह (p(x)) का शून्यक है / It is a zero of (p(x)). Since (p\(1+\sqrt{3}\)=0), \(1+\sqrt{3}\) is a zero. To prove a number is a zero, show that the polynomial value is (0).
Step 3
Exam Tip
(p\(1+\sqrt{3}\)=0), इसलिए \(1+\sqrt{3}\) शून्यक है। किसी संख्या को शून्यक सिद्ध करने के लिए बहुपद का मान (0) दिखाएँ।
The sum of zeroes is (2), so the other zero is (2-\(1+\sqrt{3}\)=1-\sqrt{3}). With rational coefficients, the conjugate also appears.
Step 2
Why this answer is correct
The correct answer is A. \(1-\sqrt{3}\). The sum of zeroes is (2), so the other zero is (2-\(1+\sqrt{3}\)=1-\sqrt{3}). With rational coefficients, the conjugate also appears.
Step 3
Exam Tip
शून्यकों का योग (2) है, इसलिए दूसरा शून्यक (2-\(1+\sqrt{3}\)=1-\sqrt{3}) है। परिमेय गुणांकों में संयुग्मी भी मिलता है।
Here \(\alpha+\beta=8\) and \(\alpha\beta=1\), so \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{64-2}{1}=62\). In such questions, first find the sum and product.
Step 2
Why this answer is correct
The correct answer is A. (62). Here \(\alpha+\beta=8\) and \(\alpha\beta=1\), so \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{64-2}{1}=62\). In such questions, first find the sum and product.
Step 3
Exam Tip
\(\alpha+\beta=8\) और \(\alpha\beta=1\), इसलिए \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{64-2}{1}=62\)। ऐसे प्रश्नों में पहले योग और गुणनफल निकालें।
