60 results found for "one-root" in Class 10.
Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(x^2+px+64=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?
If one root of \(x^2+px+64=0\) is double the other and both are negative, what is (p)?
#quadratic
#roots-relation
#parameter
A \(12\sqrt{2}\)
B \(-12\sqrt{2}\)
C \(16\sqrt{2}\)
D (24)
Explanation opens after your attempt
Correct Answer
A. \(12\sqrt{2}\)
Step 1
Concept
Let the roots be (-r) and (-2r), then \(2r^2=64\) gives \(r=4\sqrt{2}\), and \(p=3r=12\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 2
Why this answer is correct
The correct answer is A. \(12\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=64\) gives \(r=4\sqrt{2}\), and \(p=3r=12\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=64\) से \(r=4\sqrt{2}\) और \(p=3r=12\sqrt{2}\) है। परीक्षा में दोनों मूलों के चिन्ह ध्यान से रखें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(x^2-25x+q=0\) का एक मूल (10) है, तो (q) का मान क्या होगा?
If one root of \(x^2-25x+q=0\) is (10), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (150)
B (250)
C (15)
D (25)
Explanation opens after your attempt
Step 1
Concept
The other root is (15), so \(q=10\times15=150\). In exams, when (a=1), the constant term is the product of roots.
Step 2
Why this answer is correct
The correct answer is A. (150). The other root is (15), so \(q=10\times15=150\). In exams, when (a=1), the constant term is the product of roots.
Step 3
Exam Tip
दूसरा मूल (15) है, इसलिए \(q=10\times15=150\) होगा। परीक्षा में (a=1) हो तो स्थिर पद मूलों का गुणनफल होता है।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(x^2-25x+q=0\) का एक मूल (10) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-25x+q=0\) is (10), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (15)
B (10)
C (25)
D (-15)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (25), so the other root is (25-10=15). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (15). The sum of roots is (25), so the other root is (25-10=15). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (25) है, इसलिए दूसरा मूल (25-10=15) होगा। परीक्षा में एक मूल दिया हो तो योग का प्रयोग करें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35
यदि \(x^2+px+49=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?
If one root of \(x^2+px+49=0\) is double the other and both are negative, what is (p)?
#quadratic
#roots-relation
#parameter
A \( \frac{21\sqrt{2}}{2}\)
B \(-\frac{21\sqrt{2}}{2}\)
C \(14\sqrt{2}\)
D (21)
Explanation opens after your attempt
Correct Answer
A. \( \frac{21\sqrt{2}}{2}\)
Step 1
Concept
Let the roots be (-r) and (-2r), then \(2r^2=49\) and \(p=3r=\frac{21\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{21\sqrt{2}}{2}\). Let the roots be (-r) and (-2r), then \(2r^2=49\) and \(p=3r=\frac{21\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=49\) और \(p=3r=\frac{21\sqrt{2}}{2}\) है। परीक्षा में हर को परिमेय बनाना न भूलें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35
यदि \(x^2-23x+q=0\) का एक मूल (9) है, तो (q) का मान क्या होगा?
If one root of \(x^2-23x+q=0\) is (9), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (126)
B (207)
C (14)
D (23)
Explanation opens after your attempt
Step 1
Concept
The other root is (14), so \(q=9\times14=126\). In exams, when (a=1), the constant term is the product of roots.
Step 2
Why this answer is correct
The correct answer is A. (126). The other root is (14), so \(q=9\times14=126\). In exams, when (a=1), the constant term is the product of roots.
Step 3
Exam Tip
दूसरा मूल (14) है, इसलिए \(q=9\times14=126\) होगा। परीक्षा में (a=1) हो तो स्थिर पद मूलों का गुणनफल होता है।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35
यदि \(x^2-23x+q=0\) का एक मूल (9) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-23x+q=0\) is (9), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (14)
B (9)
C (23)
D (-14)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (23), so the other root is (23-9=14). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (14). The sum of roots is (23), so the other root is (23-9=14). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (23) है, इसलिए दूसरा मूल (23-9=14) होगा। परीक्षा में एक मूल दिया हो तो योग का प्रयोग करें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34
यदि \(x^2+px+36=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?
If one root of \(x^2+px+36=0\) is double the other and both are negative, what is (p)?
#quadratic
#roots-relation
#parameter
A \(9\sqrt{2}\)
B \(-9\sqrt{2}\)
C \(12\sqrt{2}\)
D (18)
Explanation opens after your attempt
Correct Answer
A. \(9\sqrt{2}\)
Step 1
Concept
Let the roots be (-r) and (-2r), then \(2r^2=36\) gives \(r=3\sqrt{2}\), and \(p=3r=9\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 2
Why this answer is correct
The correct answer is A. \(9\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=36\) gives \(r=3\sqrt{2}\), and \(p=3r=9\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=36\) से \(r=3\sqrt{2}\) और \(p=3r=9\sqrt{2}\) है। परीक्षा में दोनों मूलों के चिन्ह ध्यान से रखें।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34
यदि \(x^2-21x+q=0\) का एक मूल (8) है, तो (q) का मान क्या होगा?
If one root of \(x^2-21x+q=0\) is (8), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (104)
B (168)
C (13)
D (21)
Explanation opens after your attempt
Step 1
Concept
The other root is (13), so \(q=8\times13=104\). In exams, when (a=1), the constant term is the product of roots.
Step 2
Why this answer is correct
The correct answer is A. (104). The other root is (13), so \(q=8\times13=104\). In exams, when (a=1), the constant term is the product of roots.
Step 3
Exam Tip
दूसरा मूल (13) है, इसलिए \(q=8\times13=104\) होगा। परीक्षा में (a=1) हो तो स्थिर पद मूलों का गुणनफल होता है।
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Question
Expert Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34
यदि \(x^2-21x+q=0\) का एक मूल (8) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-21x+q=0\) is (8), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (13)
B (8)
C (21)
D (-13)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (21), so the other root is (21-8=13). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (13). The sum of roots is (21), so the other root is (21-8=13). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (21) है, इसलिए दूसरा मूल (21-8=13) होगा। परीक्षा में एक मूल दिया हो तो योग का उपयोग करें।
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Question
Hard Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(x^2+px+25=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?
If one root of \(x^2+px+25=0\) is double the other and both are negative, what is (p)?
#quadratic
#roots-relation
#parameter
A \( \frac{15\sqrt{2}}{2}\)
B \(-\frac{15\sqrt{2}}{2}\)
C \(10\sqrt{2}\)
D (15)
Explanation opens after your attempt
Correct Answer
A. \( \frac{15\sqrt{2}}{2}\)
Step 1
Concept
Let the roots be (-r) and (-2r), then \(2r^2=25\) and \(p=3r=\frac{15\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{15\sqrt{2}}{2}\). Let the roots be (-r) and (-2r), then \(2r^2=25\) and \(p=3r=\frac{15\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=25\) और \(p=3r=\frac{15\sqrt{2}}{2}\) है। परीक्षा में हर को परिमेय बनाना न भूलें।
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Question
Hard Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(x^2-15x+q=0\) का एक मूल (6) है, तो (q) का मान क्या होगा?
If one root of \(x^2-15x+q=0\) is (6), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (54)
B (90)
C (9)
D (15)
Explanation opens after your attempt
Step 1
Concept
The other root is (9), so \(q=6\times9=54\). In exams, when (a=1), (c) equals the product of roots.
Step 2
Why this answer is correct
The correct answer is A. (54). The other root is (9), so \(q=6\times9=54\). In exams, when (a=1), (c) equals the product of roots.
Step 3
Exam Tip
दूसरा मूल (9) है, इसलिए \(q=6\times9=54\) होगा। परीक्षा में (a=1) हो तो (c) मूलों के गुणनफल के बराबर होता है।
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Question
Hard Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
यदि \(x^2-15x+q=0\) का एक मूल (6) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-15x+q=0\) is (6), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (9)
B (6)
C (15)
D (-9)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (15), so the other root is (15-6=9). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (9). The sum of roots is (15), so the other root is (15-6=9). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (15) है, इसलिए दूसरा मूल (15-6=9) होगा। परीक्षा में एक मूल दिया हो तो योग का प्रयोग करें।
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Question
Hard Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35
यदि \(x^2+px+16=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?
If one root of \(x^2+px+16=0\) is double the other and both are negative, what is (p)?
#quadratic
#roots-relation
#parameter
A \(6\sqrt{2}\)
B \(-6\sqrt{2}\)
C \(8\sqrt{2}\)
D (12)
Explanation opens after your attempt
Correct Answer
A. \(6\sqrt{2}\)
Step 1
Concept
Let the roots be (-r) and (-2r), then \(2r^2=16\) gives \(r=2\sqrt{2}\), and \(p=3r=6\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 2
Why this answer is correct
The correct answer is A. \(6\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=16\) gives \(r=2\sqrt{2}\), and \(p=3r=6\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=16\) से \(r=2\sqrt{2}\) और \(p=3r=6\sqrt{2}\) है। परीक्षा में दोनों मूलों के चिन्ह ध्यान से रखें।
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Question
Hard Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35
यदि \(x^2-9x+q=0\) का एक मूल (4) है, तो (q) का मान क्या होगा?
If one root of \(x^2-9x+q=0\) is (4), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (20)
B (36)
C (5)
D (9)
Explanation opens after your attempt
Step 1
Concept
The other root is (5), so \(q=4\times5=20\). In exams, (c) equals the product of roots when (a=1).
Step 2
Why this answer is correct
The correct answer is A. (20). The other root is (5), so \(q=4\times5=20\). In exams, (c) equals the product of roots when (a=1).
Step 3
Exam Tip
दूसरा मूल (5) है, इसलिए \(q=4\times5=20\) होगा। परीक्षा में (c) मूलों के गुणनफल के बराबर होता है जब (a=1)।
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Question
Hard Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35
यदि \(x^2-9x+q=0\) का एक मूल (4) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-9x+q=0\) is (4), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (5)
B (4)
C (9)
D (-5)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (9), so the other root is (9-4=5). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (5). The sum of roots is (9), so the other root is (9-4=5). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (9) है, इसलिए दूसरा मूल (9-4=5) होगा। परीक्षा में एक मूल दिया हो तो योग का प्रयोग करें।
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Question
Hard Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34
यदि \(x^2+px+9=0\) का एक मूल दूसरे का दुगुना है और दोनों ऋणात्मक हैं, तो (p) का सही सरल मान क्या है?
If one root of \(x^2+px+9=0\) is double the other and both are negative, what is the correct simplified value of (p)?
#quadratic
#roots-relation
#simplification
A \( \frac{9\sqrt{2}}{2}\)
B \(3\sqrt{2}\)
C \(9\sqrt{2}\)
D \( \frac{3\sqrt{2}}{2}\)
Explanation opens after your attempt
Correct Answer
A. \( \frac{9\sqrt{2}}{2}\)
Step 1
Concept
\(\frac{9}{\sqrt{2}}\) simplifies to \(\frac{9\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{9\sqrt{2}}{2}\). \(\frac{9}{\sqrt{2}}\) simplifies to \(\frac{9\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 3
Exam Tip
\(\frac{9}{\sqrt{2}}\) को सरल करने पर \(\frac{9\sqrt{2}}{2}\) मिलता है। परीक्षा में हर को परिमेय बनाना न भूलें।
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Question
Hard Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34
यदि \(x^2+px+9=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?
If one root of \(x^2+px+9=0\) is double the other and both are negative, what is (p)?
#quadratic
#roots-relation
#parameter
A \(3\sqrt{2}\)
B \(-3\sqrt{2}\)
C \(9\sqrt{2}\)
D (6)
Explanation opens after your attempt
Correct Answer
A. \(3\sqrt{2}\)
Step 1
Concept
Let the roots be (-r) and (-2r), then \(2r^2=9\) and the sum is (-3r), so \(p=3r=\frac{9}{\sqrt{2}}\). In exams, assume the roots and form equations carefully.
Step 2
Why this answer is correct
The correct answer is A. \(3\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=9\) and the sum is (-3r), so \(p=3r=\frac{9}{\sqrt{2}}\). In exams, assume the roots and form equations carefully.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=9\) और योग (-3r) है, इसलिए \(p=3r=3\sqrt{\frac{9}{2}}\) नहीं बल्कि \(r=\frac{3}{\sqrt{2}}\), अतः \(p=\frac{9}{\sqrt{2}}\) होता है। परीक्षा में ऐसे प्रश्नों में मानकर समीकरण बनाएं।
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Question
Hard Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34
यदि \(x^2-5x+q=0\) का एक मूल (2) है, तो (q) का मान क्या होगा?
If one root of \(x^2-5x+q=0\) is (2), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (6)
B (10)
C (3)
D (5)
Explanation opens after your attempt
Step 1
Concept
The other root is (3), so \(q=2\times3=6\). In exams, connect (c) with product of roots.
Step 2
Why this answer is correct
The correct answer is A. (6). The other root is (3), so \(q=2\times3=6\). In exams, connect (c) with product of roots.
Step 3
Exam Tip
दूसरा मूल (3) है, इसलिए \(q=2\times3=6\) होगा। परीक्षा में (c) को मूलों के गुणनफल से जोड़ें।
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Question
Hard Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34
यदि \(x^2-5x+q=0\) का एक मूल (2) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-5x+q=0\) is (2), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (3)
B (2)
C (5)
D (-3)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (5), so the other root is (5-2=3). In exams, use sum or product when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (3). The sum of roots is (5), so the other root is (5-2=3). In exams, use sum or product when one root is given.
Step 3
Exam Tip
मूलों का योग (5) है, इसलिए दूसरा मूल (5-2=3) होगा। परीक्षा में एक मूल दिया हो तो योग या गुणनफल का प्रयोग करें।
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Question
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि (4x-2 -(4h+1)x+h=0) की एक जड़ \(\frac{1}{4}\) है, तो दूसरी जड़ क्या है?
If one root of (4x-2 -(4h+1)x+h=0) is \(\frac{1}{4}\), what is the other root?
#quadratic-roots
#other-root
#parameter
A (h)
B \(\frac{h}{4}\)
C (4h)
D (h+1)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{h}{4}\). Since one root is \(\frac{1}{4}\), the other root is (h).
Step 2
Why this answer is correct
The correct answer is A. (h). The product of roots is \(\frac{h}{4}\). Since one root is \(\frac{1}{4}\), the other root is (h).
Step 3
Exam Tip
जड़ों का गुणनफल \(\frac{h}{4}\) है। एक जड़ \(\frac{1}{4}\) है, इसलिए दूसरी जड़ (h) होगी।
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Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि (x-2 -(m+9)x+9m=0) की एक जड़ (9) है, तो दूसरी जड़ क्या है?
If one root of (x-2 -(m+9)x+9m=0) is (9), what is the other root?
#quadratic-roots
#other-root
#parameter
A (m)
B (9m)
C (m+9)
D \(\frac{m}{9}\)
Explanation opens after your attempt
Step 1
Concept
The product of roots is (9m). Since one root is (9), the other root is \(\frac{9m}{9}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (9m). Since one root is (9), the other root is \(\frac{9m}{9}=m\).
Step 3
Exam Tip
जड़ों का गुणनफल (9m) है। एक जड़ (9) है, इसलिए दूसरी जड़ \(\frac{9m}{9}=m\) होगी।
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Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि (2x-2 -(3p+2)x+p(p+2)=0) की एक जड़ (p) है, तो दूसरी जड़ क्या होगी?
If one root of (2x-2 -(3p+2)x+p(p+2)=0) is (p), what will be the other root?
#quadratic-roots
#other-root
#parametric-equation
A \(\frac{p+2}{2}\)
B (p+2)
C \(\frac{p}{2}\)
D (2p+2)
Explanation opens after your attempt
Correct Answer
A. \(\frac{p+2}{2}\)
Step 1
Concept
The product of roots is (\frac{p(p+2)}{2}). If one root is (p), the other root is \(\frac{p+2}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{p+2}{2}\). The product of roots is (\frac{p(p+2)}{2}). If one root is (p), the other root is \(\frac{p+2}{2}\).
Step 3
Exam Tip
जड़ों का गुणनफल (\frac{p(p+2)}{2}) है। एक जड़ (p) होने पर दूसरी जड़ \(\frac{p+2}{2}\) होगी।
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Question
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि (3x-2 -(3h+1)x+h=0) की एक जड़ \(\frac{1}{3}\) है, तो दूसरी जड़ क्या है?
If one root of (3x-2 -(3h+1)x+h=0) is \(\frac{1}{3}\), what is the other root?
#quadratic-roots
#other-root
#parameter
A (h)
B \(\frac{h}{3}\)
C (3h)
D (h+1)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{h}{3}\). Since one root is \(\frac{1}{3}\), the other root is (h).
Step 2
Why this answer is correct
The correct answer is A. (h). The product of roots is \(\frac{h}{3}\). Since one root is \(\frac{1}{3}\), the other root is (h).
Step 3
Exam Tip
जड़ों का गुणनफल \(\frac{h}{3}\) है। एक जड़ \(\frac{1}{3}\) है, इसलिए दूसरी जड़ (h) होगी।
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Question
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि (2x-2 -(h+1)x+h=0) की जड़ों में से एक हमेशा \(\frac{1}{2}\) है, तो दूसरी जड़ क्या होगी?
If one root of (2x-2 -(h+1)x+h=0) is always \(\frac{1}{2}\), what is the other root?
#quadratic-roots
#other-root
#parameter
A (h)
B \(\frac{h}{2}\)
C (2h)
D (h+1)
Explanation opens after your attempt
Step 1
Concept
The product is \(\frac{h}{2}\). Since one root is \(\frac{1}{2}\), the other root is \(\frac{h}{2}\div\frac{1}{2}=h\).
Step 2
Why this answer is correct
The correct answer is A. (h). The product is \(\frac{h}{2}\). Since one root is \(\frac{1}{2}\), the other root is \(\frac{h}{2}\div\frac{1}{2}=h\).
Step 3
Exam Tip
गुणनफल \(\frac{h}{2}\) है। एक जड़ \(\frac{1}{2}\) है, इसलिए दूसरी जड़ \(\frac{h}{2}\div\frac{1}{2}=h\) होगी।
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Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि (x-2 -(m+2)x+3m=0) की एक जड़ (3) है, तो दूसरी जड़ क्या है?
If one root of (x-2 -(m+2)x+3m=0) is (3), what is the other root?
#quadratic-roots
#other-root
#parametric-equation
A (m)
B (3m)
C (m+2)
D \(\frac{m}{3}\)
Explanation opens after your attempt
Step 1
Concept
Putting (x=3) makes the equation true for every (m). The product is (3m) and one root is (3), so the other root is (m).
Step 2
Why this answer is correct
The correct answer is A. (m). Putting (x=3) makes the equation true for every (m). The product is (3m) and one root is (3), so the other root is (m).
Step 3
Exam Tip
(x=3) रखने पर समीकरण हर (m) के लिए सही हो जाता है। गुणनफल (3m) है और एक जड़ (3), इसलिए दूसरी जड़ (m) है।
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Question
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि (x-2 -(m-2 )x+m-6 =0) की एक जड़ (3) है, तो दूसरी जड़ क्या होगी?
If one root of (x-2 -(m-2 )x+m-6 =0) is (3), what is the other root?
#quadratic-roots
#other-root
#error-check
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
Putting (x=3) gives (9-3(m-2 )+m-6 =0), so \(m=\frac{9}{2}\). The product is \(-\frac{3}{2}\), so the other root is \(-\frac{1}{2}\); hence no option is correct.
Step 2
Why this answer is correct
The correct answer is A. (1). Putting (x=3) gives (9-3(m-2 )+m-6 =0), so \(m=\frac{9}{2}\). The product is \(-\frac{3}{2}\), so the other root is \(-\frac{1}{2}\); hence no option is correct.
Step 3
Exam Tip
(x=3) रखने पर (9-3(m-2 )+m-6 =0), इसलिए \(m=\frac{9}{2}\)। गुणनफल \(m-6=-\frac{3}{2}\) है, अतः दूसरी जड़ \(-\frac{1}{2}\) होगी, इसलिए कोई विकल्प सही नहीं है।
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Question
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि \(x^2+ax+12=0\) की एक जड़ दूसरी जड़ से (3) अधिक है, तो (a) के संभव मान क्या हैं?
If one root of \(x^2+ax+12=0\) is (3) more than the other root, what are the possible values of (a)?
#quadratic-roots
#difference-of-roots
#parameter
A \(\sqrt{57}\) और \(-\sqrt{57}\) / \(\sqrt{57}\) and \(-\sqrt{57}\)
B \(\sqrt{21}\) और \(-\sqrt{21}\) / \(\sqrt{21}\) and \(-\sqrt{21}\)
C (3) और (-3) / (3) and (-3)
D (12) और (-12) / (12) and (-12)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{57}\) और \(-\sqrt{57}\) / \(\sqrt{57}\) and \(-\sqrt{57}\)
Step 1
Concept
Let the roots be (r) and (r+3). Then (r(r+3)=12), giving the sum as \(\pm\sqrt{57}\), so \(a=\mp\sqrt{57}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{57}\) और \(-\sqrt{57}\) / \(\sqrt{57}\) and \(-\sqrt{57}\). Let the roots be (r) and (r+3). Then (r(r+3)=12), giving the sum as \(\pm\sqrt{57}\), so \(a=\mp\sqrt{57}\).
Step 3
Exam Tip
जड़ें (r) और (r+3) मानने पर (r(r+3)=12) मिलता है। इससे जड़ों का योग \(\pm\sqrt{57}\) होता है, इसलिए \(a=\mp\sqrt{57}\)।
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Question
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि (x-2 +(k-3)x+k=0) की एक जड़ दूसरी जड़ की दुगुनी है, तो (k) का मान क्या होगा?
If one root of (x-2 +(k-3)x+k=0) is twice the other root, what is the value of (k)?
#quadratic-roots
#roots-ratio
#parameter
A \(\frac{21+3\sqrt{33}}{4}\) या \(\frac{21-3\sqrt{33}}{4}\) / \(\frac{21+3\sqrt{33}}{4}\) or \(\frac{21-3\sqrt{33}}{4}\)
B \(\frac{3+\sqrt{33}}{4}\) या \(\frac{3-\sqrt{33}}{4}\) / \(\frac{3+\sqrt{33}}{4}\) or \(\frac{3-\sqrt{33}}{4}\)
C (6) या (3) / (6) or (3)
D (9) या (2) / (9) or (2)
Explanation opens after your attempt
Correct Answer
A. \(\frac{21+3\sqrt{33}}{4}\) या \(\frac{21-3\sqrt{33}}{4}\) / \(\frac{21+3\sqrt{33}}{4}\) or \(\frac{21-3\sqrt{33}}{4}\)
Step 1
Concept
Taking the roots as (r) and (2r), we get (3r=3-k) and \(2r^2=k\). Solving \(2k^2-21k+18=0\) gives the two listed values.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{21+3\sqrt{33}}{4}\) या \(\frac{21-3\sqrt{33}}{4}\) / \(\frac{21+3\sqrt{33}}{4}\) or \(\frac{21-3\sqrt{33}}{4}\). Taking the roots as (r) and (2r), we get (3r=3-k) and \(2r^2=k\). Solving \(2k^2-21k+18=0\) gives the two listed values.
Step 3
Exam Tip
जड़ें (r) और (2r) मानने पर (3r=3-k) और \(2r^2=k\) मिलता है। हल करने पर \(2k^2-21k+18=0\), इसलिए दिए गए दोनों मान मिलते हैं।
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Question
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि \(x^2+ax+a=0\) का एक मूल (2) है तो दूसरा मूल क्या होगा?
If one root of \(x^2+ax+a=0\) is (2), what will be the other root?
#roots
#parameter
#other_root
A \(-\frac{2}{3}\)
B \(\frac{2}{3}\)
C (3)
D (-3)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{2}{3}\)
Step 1
Concept
Putting (x=2) gives (4+3a=0), so \(a=-\frac{4}{3}\). The product is (a), so the other root is \(-\frac{2}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{2}{3}\). Putting (x=2) gives (4+3a=0), so \(a=-\frac{4}{3}\). The product is (a), so the other root is \(-\frac{2}{3}\).
Step 3
Exam Tip
(x=2) रखने पर (4+3a=0) से \(a=-\frac{4}{3}\) है। गुणनफल (a) है इसलिए दूसरा मूल \(-\frac{2}{3}\) होगा।
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Question
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि \(x^2+ax+a=0\) का एक मूल (1) है तो दूसरा मूल क्या होगा?
If one root of \(x^2+ax+a=0\) is (1), what will be the other root?
#roots
#parameter
#other_root
A \(-\frac{1}{2}\)
B \(\frac{1}{2}\)
C (2)
D (-2)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{1}{2}\)
Step 1
Concept
Putting (x=1) gives (1+2a=0), so \(a=-\frac{1}{2}\). The product is (a), so the other root is \(-\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{1}{2}\). Putting (x=1) gives (1+2a=0), so \(a=-\frac{1}{2}\). The product is (a), so the other root is \(-\frac{1}{2}\).
Step 3
Exam Tip
(x=1) रखने पर (1+2a=0) से \(a=-\frac{1}{2}\) है। गुणनफल (a) है इसलिए दूसरा मूल \(-\frac{1}{2}\) होगा।
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Question
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
समीकरण \(x^2-17x+70=0\) का एक मूल (7) है तो दूसरा मूल क्या है?
If one root of \(x^2-17x+70=0\) is (7), what is the other root?
#roots
#other_root
#product
A (10)
B (7)
C (17)
D (70)
Explanation opens after your attempt
Step 1
Concept
The product of roots is (70) and one root is (7). The other root is \(\frac{70}{7}=10\).
Step 2
Why this answer is correct
The correct answer is A. (10). The product of roots is (70) and one root is (7). The other root is \(\frac{70}{7}=10\).
Step 3
Exam Tip
मूलों का गुणनफल (70) है और एक मूल (7) है। दूसरा मूल \(\frac{70}{7}=10\) होगा।
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Question
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
समीकरण \(x^2-15x+54=0\) का एक मूल (6) है तो दूसरा मूल क्या है?
If one root of \(x^2-15x+54=0\) is (6), what is the other root?
#roots
#other_root
#product
A (9)
B (6)
C (15)
D (54)
Explanation opens after your attempt
Step 1
Concept
The product of roots is (54) and one root is (6). Therefore the other root is \(\frac{54}{6}=9\).
Step 2
Why this answer is correct
The correct answer is A. (9). The product of roots is (54) and one root is (6). Therefore the other root is \(\frac{54}{6}=9\).
Step 3
Exam Tip
मूलों का गुणनफल (54) है और एक मूल (6) है। इसलिए दूसरा मूल \(\frac{54}{6}=9\) है।
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Question
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
समीकरण \(x^2-13x+40=0\) का एक मूल (5) है तो दूसरा मूल क्या है?
If one root of \(x^2-13x+40=0\) is (5), what is the other root?
#roots
#other_root
#product
A (8)
B (5)
C (13)
D (40)
Explanation opens after your attempt
Step 1
Concept
The product of roots is (40) and one root is (5). Therefore the other root is \(\frac{40}{5}=8\).
Step 2
Why this answer is correct
The correct answer is A. (8). The product of roots is (40) and one root is (5). Therefore the other root is \(\frac{40}{5}=8\).
Step 3
Exam Tip
मूलों का गुणनफल (40) है और एक मूल (5) है। इसलिए दूसरा मूल \(\frac{40}{5}=8\) है।
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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि मूलों का योग (12) है और एक मूल (5) है तो दूसरा मूल क्या है?
If the sum of roots is (12) and one root is (5), what is the other root?
#roots
#other_root
#sum
A (5)
B (7)
C (12)
D (17)
Explanation opens after your attempt
Step 1
Concept
The other root is (12-5=7). Subtract the given root from the sum.
Step 2
Why this answer is correct
The correct answer is B. (7). The other root is (12-5=7). Subtract the given root from the sum.
Step 3
Exam Tip
दूसरा मूल (12-5=7) है। योग में से दिया हुआ मूल घटाएं।
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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि एक मूल (6) है और मूलों का गुणनफल (48) है तो दूसरा मूल क्या होगा?
If one root is (6) and the product of roots is (48), what is the other root?
#roots
#other_root
#product
A (6)
B (8)
C (42)
D (48)
Explanation opens after your attempt
Step 1
Concept
The other root is \(\frac{48}{6}=8\). Divide the product by the given root.
Step 2
Why this answer is correct
The correct answer is B. (8). The other root is \(\frac{48}{6}=8\). Divide the product by the given root.
Step 3
Exam Tip
दूसरा मूल \(\frac{48}{6}=8\) होगा। गुणनफल को दिए हुए मूल से भाग करें।
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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
समीकरण \(x^2-11x+30=0\) का एक मूल (5) है तो दूसरा मूल क्या है?
If one root of \(x^2-11x+30=0\) is (5), what is the other root?
#roots
#other_root
#factorisation
A (5)
B (6)
C (11)
D (30)
Explanation opens after your attempt
Step 1
Concept
(x-2 -11x+30=(x-5)(x-6)). Therefore the other root is (6).
Step 2
Why this answer is correct
The correct answer is B. (6). (x-2 -11x+30=(x-5)(x-6)). Therefore the other root is (6).
Step 3
Exam Tip
(x-2 -11x+30=(x-5)(x-6)) है। इसलिए दूसरा मूल (6) है।
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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि मूलों का योग (8) है और एक मूल (3) है तो दूसरा मूल क्या है?
If the sum of roots is (8) and one root is (3), what is the other root?
#roots
#other_root
#sum
A (3)
B (5)
C (8)
D (11)
Explanation opens after your attempt
Step 1
Concept
The other root is (8-3=5). Subtract the given root from the sum.
Step 2
Why this answer is correct
The correct answer is B. (5). The other root is (8-3=5). Subtract the given root from the sum.
Step 3
Exam Tip
दूसरा मूल (8-3=5) है। योग में से दिया हुआ मूल घटाएं।
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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि एक मूल (5) है और मूलों का गुणनफल (35) है तो दूसरा मूल क्या होगा?
If one root is (5) and the product of roots is (35), what is the other root?
#roots
#other_root
#product
A (5)
B (7)
C (30)
D (35)
Explanation opens after your attempt
Step 1
Concept
The other root is \(\frac{35}{5}=7\). Divide the product by the given root.
Step 2
Why this answer is correct
The correct answer is B. (7). The other root is \(\frac{35}{5}=7\). Divide the product by the given root.
Step 3
Exam Tip
दूसरा मूल \(\frac{35}{5}=7\) होगा। गुणनफल में दिए हुए मूल से भाग करें।
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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
समीकरण \(x^2-9x+18=0\) का एक मूल (3) है तो दूसरा मूल क्या है?
If one root of \(x^2-9x+18=0\) is (3), what is the other root?
#roots
#other_root
#factorisation
A (3)
B (6)
C (9)
D (18)
Explanation opens after your attempt
Step 1
Concept
(x-2 -9x+18=(x-3)(x-6)). Therefore the other root is (6).
Step 2
Why this answer is correct
The correct answer is B. (6). (x-2 -9x+18=(x-3)(x-6)). Therefore the other root is (6).
Step 3
Exam Tip
(x-2 -9x+18=(x-3)(x-6)) है। इसलिए दूसरा मूल (6) है।
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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
निम्न में से किस समीकरण का एक मूल (0) है?
Which of the following equations has one root (0)?
#roots
#zero_root
#identification
A \(x^2+5x=0\)
B \(x^2+5=0\)
C \(x^2-5x+6=0\)
D \(x^2+2x+1=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+5x=0\)
Step 1
Concept
(x-2 +5x=x(x+5)) so (x=0) is one root. If the constant term is (0) check for a zero root.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+5x=0\). (x-2 +5x=x(x+5)) so (x=0) is one root. If the constant term is (0) check for a zero root.
Step 3
Exam Tip
(x-2 +5x=x(x+5)) इसलिए (x=0) एक मूल है। अचर पद (0) हो तो शून्य मूल की संभावना देखें।
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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि मूलों का योग (5) है और एक मूल (2) है तो दूसरा मूल क्या है?
If the sum of roots is (5) and one root is (2) then what is the other root?
#roots
#other_root
#sum
A (2)
B (3)
C (5)
D (7)
Explanation opens after your attempt
Step 1
Concept
The other root is (5-2=3). Subtract the given root from the sum.
Step 2
Why this answer is correct
The correct answer is B. (3). The other root is (5-2=3). Subtract the given root from the sum.
Step 3
Exam Tip
दूसरा मूल (5-2=3) है। योग में से दिए हुए मूल को घटाएं।
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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि एक मूल (3) है और मूलों का गुणनफल (12) है तो दूसरा मूल क्या होगा?
If one root is (3) and the product of roots is (12) then what is the other root?
#roots
#other_root
#product
A (3)
B (4)
C (9)
D (12)
Explanation opens after your attempt
Step 1
Concept
The other root is \(\frac{12}{3}=4\). In product questions divide by the given root.
Step 2
Why this answer is correct
The correct answer is B. (4). The other root is \(\frac{12}{3}=4\). In product questions divide by the given root.
Step 3
Exam Tip
दूसरा मूल \(\frac{12}{3}=4\) होगा। गुणनफल वाले प्रश्न में दिए मूल से भाग दें।
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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
समीकरण \(x^2-6x+8=0\) का एक मूल (4) है तो दूसरा मूल क्या है?
If one root of \(x^2-6x+8=0\) is (4) then what is the other root?
#roots
#other_root
#factorisation
A (1)
B (2)
C (-2)
D (8)
Explanation opens after your attempt
Step 1
Concept
(x-2 -6x+8=(x-4)(x-2)) so the other root is (2). Use the given root to find the other factor.
Step 2
Why this answer is correct
The correct answer is B. (2). (x-2 -6x+8=(x-4)(x-2)) so the other root is (2). Use the given root to find the other factor.
Step 3
Exam Tip
(x-2 -6x+8=(x-4)(x-2)) इसलिए दूसरा मूल (2) है। दिए गए एक मूल से दूसरा गुणनखंड खोजें।
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Question
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(x^2+mx+64=0\) का एक मूल दूसरे का व्युत्क्रम है, तो (m) के बारे में क्या कहा जा सकता है?
If one root of \(x^2+mx+64=0\) is the reciprocal of the other, what can be said about (m)?
#quadratic-equations
#reciprocal-roots
#conceptual
#expert
A ऐसा संभव नहीं है / It is not possible
B (m=64)
C (m=-64)
D (m=0)
Explanation opens after your attempt
Correct Answer
A. ऐसा संभव नहीं है / It is not possible
Step 1
Concept
If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (64), so it is not possible.
Step 2
Why this answer is correct
The correct answer is A. ऐसा संभव नहीं है / It is not possible. If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (64), so it is not possible.
Step 3
Exam Tip
एक मूल दूसरे का व्युत्क्रम हो तो मूलों का गुणनफल (1) होना चाहिए। यहाँ गुणनफल (64) है, इसलिए ऐसा संभव नहीं है।
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Question
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि (x-2 -(m+7)x+7m=0) का एक मूल (7) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+7)x+7m=0) is (7), what is the other root?
#quadratic-equations
#one-root
#roots
#expert
A (m)
B (7m)
C (m+7)
D (m-7 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (7m) and one root is (7). Hence the other root is \(\frac{7m}{7}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (7m) and one root is (7). Hence the other root is \(\frac{7m}{7}=m\).
Step 3
Exam Tip
मूलों का गुणनफल (7m) है और एक मूल (7) है। इसलिए दूसरा मूल \(\frac{7m}{7}=m\) है।
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Question
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(x^2+ax+54=0\) का एक मूल (6) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+54=0\) is (6), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#expert
A दूसरा मूल (9), (a=-15) / other root (9), (a=-15)
B दूसरा मूल (-9), (a=3) / other root (-9), (a=3)
C दूसरा मूल (9), (a=15) / other root (9), (a=15)
D दूसरा मूल (-6), (a=0) / other root (-6), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (9), (a=-15) / other root (9), (a=-15)
Step 1
Concept
The product of roots is (54), so the other root is (9). The sum is (15), and (-a=15), so (a=-15).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (9), (a=-15) / other root (9), (a=-15). The product of roots is (54), so the other root is (9). The sum is (15), and (-a=15), so (a=-15).
Step 3
Exam Tip
मूलों का गुणनफल (54) है, इसलिए दूसरा मूल (9) होगा। योग (15) है और (-a=15), इसलिए (a=-15)।
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Question
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(x^2+mx+49=0\) का एक मूल दूसरे का व्युत्क्रम है, तो (m) के बारे में क्या कहा जा सकता है?
If one root of \(x^2+mx+49=0\) is the reciprocal of the other, what can be said about (m)?
#quadratic-equations
#reciprocal-roots
#conceptual
#expert
A ऐसा संभव नहीं है / It is not possible
B (m=49)
C (m=-49)
D (m=0)
Explanation opens after your attempt
Correct Answer
A. ऐसा संभव नहीं है / It is not possible
Step 1
Concept
If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (49), so it is not possible.
Step 2
Why this answer is correct
The correct answer is A. ऐसा संभव नहीं है / It is not possible. If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (49), so it is not possible.
Step 3
Exam Tip
एक मूल दूसरे का व्युत्क्रम हो तो मूलों का गुणनफल (1) होना चाहिए। यहाँ गुणनफल (49) है, इसलिए ऐसा संभव नहीं है।
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Question
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि (x-2 -(m+6)x+6m=0) का एक मूल (6) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+6)x+6m=0) is (6), what is the other root?
#quadratic-equations
#one-root
#roots
#expert
A (m)
B (6m)
C (m+6)
D (m-6 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (6m) and one root is (6). Hence the other root is \(\frac{6m}{6}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (6m) and one root is (6). Hence the other root is \(\frac{6m}{6}=m\).
Step 3
Exam Tip
मूलों का गुणनफल (6m) है और एक मूल (6) है। इसलिए दूसरा मूल \(\frac{6m}{6}=m\) है।
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Question
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(x^2+ax+40=0\) का एक मूल (5) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+40=0\) is (5), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#expert
A दूसरा मूल (8), (a=-13) / other root (8), (a=-13)
B दूसरा मूल (-8), (a=3) / other root (-8), (a=3)
C दूसरा मूल (8), (a=13) / other root (8), (a=13)
D दूसरा मूल (-5), (a=0) / other root (-5), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (8), (a=-13) / other root (8), (a=-13)
Step 1
Concept
The product of roots is (40), so the other root is (8). The sum is (13), and (-a=13), so (a=-13).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (8), (a=-13) / other root (8), (a=-13). The product of roots is (40), so the other root is (8). The sum is (13), and (-a=13), so (a=-13).
Step 3
Exam Tip
मूलों का गुणनफल (40) है, इसलिए दूसरा मूल (8) होगा। योग (13) है और (-a=13), इसलिए (a=-13)।
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Question
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 28
यदि \(x^2+mx+36=0\) का एक मूल दूसरे का व्युत्क्रम है, तो (m) के बारे में क्या कहा जा सकता है?
If one root of \(x^2+mx+36=0\) is the reciprocal of the other, what can be said about (m)?
#quadratic-equations
#reciprocal-roots
#conceptual
#expert
A ऐसा संभव नहीं है / It is not possible
B (m=36)
C (m=-36)
D (m=0)
Explanation opens after your attempt
Correct Answer
A. ऐसा संभव नहीं है / It is not possible
Step 1
Concept
If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (36), so it is not possible.
Step 2
Why this answer is correct
The correct answer is A. ऐसा संभव नहीं है / It is not possible. If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (36), so it is not possible.
Step 3
Exam Tip
एक मूल दूसरे का व्युत्क्रम हो तो मूलों का गुणनफल (1) होना चाहिए। यहाँ गुणनफल (36) है, इसलिए ऐसा संभव नहीं है।
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Question
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 28
यदि (x-2 -(m+5)x+5m=0) का एक मूल (5) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+5)x+5m=0) is (5), what is the other root?
#quadratic-equations
#one-root
#roots
#expert
A (m)
B (5m)
C (m+5)
D (m-5 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (5m) and one root is (5). Hence the other root is \(\frac{5m}{5}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (5m) and one root is (5). Hence the other root is \(\frac{5m}{5}=m\).
Step 3
Exam Tip
मूलों का गुणनफल (5m) है और एक मूल (5) है। इसलिए दूसरा मूल \(\frac{5m}{5}=m\) है।
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Question
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 28
यदि \(x^2+ax+24=0\) का एक मूल (4) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+24=0\) is (4), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#expert
A दूसरा मूल (6), (a=-10) / other root (6), (a=-10)
B दूसरा मूल (-6), (a=2) / other root (-6), (a=2)
C दूसरा मूल (6), (a=10) / other root (6), (a=10)
D दूसरा मूल (-4), (a=0) / other root (-4), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (6), (a=-10) / other root (6), (a=-10)
Step 1
Concept
The product of roots is (24), so the other root is (6). The sum is (10), and (-a=10), so (a=-10).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (6), (a=-10) / other root (6), (a=-10). The product of roots is (24), so the other root is (6). The sum is (10), and (-a=10), so (a=-10).
Step 3
Exam Tip
मूलों का गुणनफल (24) है, इसलिए दूसरा मूल (6) होगा। योग (10) है और (-a=10), इसलिए (a=-10)।
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Question
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
किस समीकरण में (x=0) एक मूल है और दूसरा मूल ऋणात्मक है?
In which equation is (x=0) one root and the other root negative?
#quadratic-equations
#zero-root
#root-sign
#hard
A \(x^2+7x=0\)
B \(x^2-7x=0\)
C \(x^2+7=0\)
D \(x^2-7=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+7x=0\)
Step 1
Concept
(x-2 +7x=x(x+7)), so the roots are (0) and (-7). The other root is negative.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+7x=0\). (x-2 +7x=x(x+7)), so the roots are (0) and (-7). The other root is negative.
Step 3
Exam Tip
(x-2 +7x=x(x+7)), इसलिए मूल (0) और (-7) हैं। दूसरा मूल ऋणात्मक है।
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Question
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(x^2+mx+25=0\) का एक मूल दूसरे का व्युत्क्रम है, तो (m) के बारे में क्या कहा जा सकता है?
If one root of \(x^2+mx+25=0\) is the reciprocal of the other, what can be said about (m)?
#quadratic-equations
#reciprocal-roots
#conceptual
#hard
A ऐसा संभव नहीं है / It is not possible
B (m=25)
C (m=-25)
D (m=0)
Explanation opens after your attempt
Correct Answer
A. ऐसा संभव नहीं है / It is not possible
Step 1
Concept
If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (25), so it is not possible.
Step 2
Why this answer is correct
The correct answer is A. ऐसा संभव नहीं है / It is not possible. If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (25), so it is not possible.
Step 3
Exam Tip
एक मूल दूसरे का व्युत्क्रम हो तो मूलों का गुणनफल (1) होना चाहिए। यहाँ गुणनफल (25) है, इसलिए ऐसा संभव नहीं है।
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Question
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि (x-2 -(m+4)x+4m=0) का एक मूल (4) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+4)x+4m=0) is (4), what is the other root?
#quadratic-equations
#one-root
#roots
#hard
A (m)
B (4m)
C (m+4)
D (m-4 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (4m) and one root is (4). Hence the other root is \(\frac{4m}{4}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (4m) and one root is (4). Hence the other root is \(\frac{4m}{4}=m\).
Step 3
Exam Tip
गुणनफल (4m) है और एक मूल (4) है। इसलिए दूसरा मूल \(\frac{4m}{4}=m\) है।
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Question
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(x^2+ax+18=0\) का एक मूल (2) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+18=0\) is (2), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#hard
A दूसरा मूल (9), (a=-11) / other root (9), (a=-11)
B दूसरा मूल (-9), (a=7) / other root (-9), (a=7)
C दूसरा मूल (9), (a=11) / other root (9), (a=11)
D दूसरा मूल (-2), (a=0) / other root (-2), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (9), (a=-11) / other root (9), (a=-11)
Step 1
Concept
The product of roots is (18), so the other root is (9). The sum is (11), and (-a=11), so (a=-11).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (9), (a=-11) / other root (9), (a=-11). The product of roots is (18), so the other root is (9). The sum is (11), and (-a=11), so (a=-11).
Step 3
Exam Tip
मूलों का गुणनफल (18) है, इसलिए दूसरा मूल (9) होगा। योग (11) है और (-a=11), इसलिए (a=-11)।
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Question
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
किस समीकरण में (x=0) एक मूल है और दूसरा मूल धनात्मक है?
In which equation is (x=0) one root and the other root positive?
#quadratic-equations
#zero-root
#root-sign
#hard
A \(x^2-6x=0\)
B \(x^2+6x=0\)
C \(x^2+6=0\)
D \(x^2-6=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-6x=0\)
Step 1
Concept
(x-2 -6x=x(x-6)), so the roots are (0) and (6). The other root is positive.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x=0\). (x-2 -6x=x(x-6)), so the roots are (0) and (6). The other root is positive.
Step 3
Exam Tip
(x-2 -6x=x(x-6)), इसलिए मूल (0) और (6) हैं। दूसरा मूल धनात्मक है।
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Question
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(x^2+mx+16=0\) का एक मूल दूसरे का व्युत्क्रम है, तो (m) के बारे में क्या कहा जा सकता है?
If one root of \(x^2+mx+16=0\) is the reciprocal of the other, what can be said about (m)?
#quadratic-equations
#reciprocal-roots
#conceptual
#hard
A ऐसा संभव नहीं है / It is not possible
B (m=16)
C (m=-16)
D (m=0)
Explanation opens after your attempt
Correct Answer
A. ऐसा संभव नहीं है / It is not possible
Step 1
Concept
If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (16), so it is not possible.
Step 2
Why this answer is correct
The correct answer is A. ऐसा संभव नहीं है / It is not possible. If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (16), so it is not possible.
Step 3
Exam Tip
एक मूल दूसरे का व्युत्क्रम हो तो मूलों का गुणनफल (1) होना चाहिए। यहाँ गुणनफल (16) है, इसलिए ऐसा संभव नहीं है।
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Question
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि (x-2 -(m+2)x+2m=0) का एक मूल (2) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+2)x+2m=0) is (2), what is the other root?
#quadratic-equations
#one-root
#roots
#hard
A (m)
B (2m)
C (m+2)
D (m-2 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (2m) and one root is (2). Hence the other root is \(\frac{2m}{2}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (2m) and one root is (2). Hence the other root is \(\frac{2m}{2}=m\).
Step 3
Exam Tip
गुणनफल (2m) है और एक मूल (2) है। इसलिए दूसरा मूल \(\frac{2m}{2}=m\) है।
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Question
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(x^2+ax+12=0\) का एक मूल (3) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+12=0\) is (3), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#hard
A दूसरा मूल (4), (a=-7) / other root (4), (a=-7)
B दूसरा मूल (-4), (a=1) / other root (-4), (a=1)
C दूसरा मूल (4), (a=7) / other root (4), (a=7)
D दूसरा मूल (-3), (a=0) / other root (-3), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (4), (a=-7) / other root (4), (a=-7)
Step 1
Concept
The product of roots is (12), so the other root is (4). The sum is (7), and (-a=7), so (a=-7).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (4), (a=-7) / other root (4), (a=-7). The product of roots is (12), so the other root is (4). The sum is (7), and (-a=7), so (a=-7).
Step 3
Exam Tip
मूलों का गुणनफल (12) है, इसलिए दूसरा मूल (4) होगा। योग (7) है और (-a=7), इसलिए (a=-7)।
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