12 results found for "monic-equation" in Class 10.
Question
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि \(x^2-6x-16=0\) के मूलों को (1) बढ़ा दिया जाए तो नए मूलों से बना मोनिक समीकरण कौन सा होगा?
If each root of \(x^2-6x-16=0\) is increased by (1), which monic equation is formed from the new roots?
Explanation opens after your attempt
Correct Answer
A. \(x^2-8x-9=0\)
Step 1
Concept
The old roots are (8) and (-2). The new roots are (9) and (-1), so the equation is \(x^2-8x-9=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-8x-9=0\). The old roots are (8) and (-2). The new roots are (9) and (-1), so the equation is \(x^2-8x-9=0\).
Step 3
Exam Tip
पुराने मूल (8) और (-2) हैं। नए मूल (9) और (-1) होंगे इसलिए समीकरण \(x^2-8x-9=0\) है।
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Question
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि \(x^2-4x-12=0\) के मूलों को (1) बढ़ा दिया जाए तो नए मूलों से बना मोनिक समीकरण कौन सा होगा?
If each root of \(x^2-4x-12=0\) is increased by (1), which monic equation is formed from the new roots?
Explanation opens after your attempt
Correct Answer
A. \(x^2-6x-15=0\)
Step 1
Concept
The old roots are (6) and (-2). The new roots are (7) and (-1), so the equation is \(x^2-6x-7=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x-15=0\). The old roots are (6) and (-2). The new roots are (7) and (-1), so the equation is \(x^2-6x-7=0\).
Step 3
Exam Tip
पुराने मूल (6) और (-2) हैं। नए मूल (7) और (-1) होंगे इसलिए समीकरण \(x^2-6x-7=0\) नहीं बल्कि \(x^2-6x-7=0\) होता है।
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Question
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि \(\alpha+\beta=-7\) और \(\alpha\beta=-18\) है तो \(\alpha\) और \(\beta\) के लिए मोनिक समीकरण कौन सा है?
If \(\alpha+\beta=-7\) and \(\alpha\beta=-18\), which monic equation has roots \(\alpha\) and \(\beta\)?
Explanation opens after your attempt
Correct Answer
A. \(x^2+7x-18=0\)
Step 1
Concept
The monic equation is (x-2-\(\alpha+\beta\)x+\alpha\beta=0). Therefore \(x^2+7x-18=0\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+7x-18=0\). The monic equation is (x-2-\(\alpha+\beta\)x+\alpha\beta=0). Therefore \(x^2+7x-18=0\) is correct.
Step 3
Exam Tip
मोनिक समीकरण (x-2-\(\alpha+\beta\)x+\alpha\beta=0) होता है। इसलिए \(x^2+7x-18=0\) सही है।
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Question
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि मूलों का योग (0) और गुणनफल (-36) है तो मोनिक समीकरण कौन सा होगा?
If the sum of roots is (0) and product is (-36), which monic equation is formed?
Explanation opens after your attempt
Correct Answer
A. \(x^2-36=0\)
Step 1
Concept
The monic equation is (x-2-(0)x+(-36)=0). Therefore \(x^2-36=0\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-36=0\). The monic equation is (x-2-(0)x+(-36)=0). Therefore \(x^2-36=0\) is correct.
Step 3
Exam Tip
मोनिक समीकरण (x-2-(0)x+(-36)=0) होगा। इसलिए \(x^2-36=0\) सही है।
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Question
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि \(\alpha+\beta=-5\) और \(\alpha\beta=-14\) है तो \(\alpha\) और \(\beta\) के लिए मोनिक समीकरण कौन सा है?
If \(\alpha+\beta=-5\) and \(\alpha\beta=-14\), which monic equation has roots \(\alpha\) and \(\beta\)?
Explanation opens after your attempt
Correct Answer
A. \(x^2+5x-14=0\)
Step 1
Concept
The monic equation is (x-2-\(\alpha+\beta\)x+\alpha\beta=0). Therefore \(x^2+5x-14=0\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+5x-14=0\). The monic equation is (x-2-\(\alpha+\beta\)x+\alpha\beta=0). Therefore \(x^2+5x-14=0\) is correct.
Step 3
Exam Tip
मोनिक समीकरण (x-2-\(\alpha+\beta\)x+\alpha\beta=0) होता है। इसलिए \(x^2+5x-14=0\) सही है।
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Question
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि मूलों का योग (0) और गुणनफल (-25) है तो मोनिक समीकरण कौन सा होगा?
If the sum of roots is (0) and product is (-25), which monic equation is formed?
Explanation opens after your attempt
Correct Answer
A. \(x^2-25=0\)
Step 1
Concept
The monic equation is (x-2-(0)x+(-25)=0). Therefore \(x^2-25=0\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-25=0\). The monic equation is (x-2-(0)x+(-25)=0). Therefore \(x^2-25=0\) is correct.
Step 3
Exam Tip
मोनिक समीकरण (x-2-(0)x+(-25)=0) होगा। इसलिए \(x^2-25=0\) सही है।
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Question
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि \(\alpha+\beta=-3\) और \(\alpha\beta=-10\) है तो \(\alpha\) और \(\beta\) के लिए मोनिक समीकरण कौन सा है?
If \(\alpha+\beta=-3\) and \(\alpha\beta=-10\), which monic equation has roots \(\alpha\) and \(\beta\)?
Explanation opens after your attempt
Correct Answer
A. \(x^2+3x-10=0\)
Step 1
Concept
The monic equation is (x-2-\(\alpha+\beta\)x+\alpha\beta=0). Therefore \(x^2+3x-10=0\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+3x-10=0\). The monic equation is (x-2-\(\alpha+\beta\)x+\alpha\beta=0). Therefore \(x^2+3x-10=0\) is correct.
Step 3
Exam Tip
मोनिक समीकरण (x-2-\(\alpha+\beta\)x+\alpha\beta=0) होता है। इसलिए \(x^2+3x-10=0\) सही है।
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Question
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि मूलों का योग (0) और गुणनफल (-16) है तो मोनिक समीकरण कौन सा होगा?
If the sum of roots is (0) and product is (-16), which monic equation is formed?
Explanation opens after your attempt
Correct Answer
A. \(x^2-16=0\)
Step 1
Concept
The monic equation is (x-2-(0)x+(-16)=0). Therefore \(x^2-16=0\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-16=0\). The monic equation is (x-2-(0)x+(-16)=0). Therefore \(x^2-16=0\) is correct.
Step 3
Exam Tip
मोनिक समीकरण (x-2-(0)x+(-16)=0) होगा। इसलिए \(x^2-16=0\) सही है।
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Question
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि किसी मोनिक द्विघात समीकरण के मूल (3r) और (4r) हैं तथा उनका योग (28) है, तो उस समीकरण का स्थिर पद क्या होगा?
If the roots of a monic quadratic equation are (3r) and (4r), and their sum is (28), what will be the constant term?
Explanation opens after your attempt
Step 1
Concept
From (3r+4r=28), we get (r=4), so the roots are (12) and (16). The constant term is the product of roots (192).
Step 2
Why this answer is correct
The correct answer is A. (192). From (3r+4r=28), we get (r=4), so the roots are (12) and (16). The constant term is the product of roots (192).
Step 3
Exam Tip
(3r+4r=28) से (r=4) मिलता है, इसलिए मूल (12) और (16) हैं। स्थिर पद मूलों का गुणनफल (192) होगा।
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Question
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि किसी मोनिक द्विघात समीकरण के मूल (2r) और (5r) हैं तथा उनका योग (21) है, तो उस समीकरण का स्थिर पद क्या होगा?
If the roots of a monic quadratic equation are (2r) and (5r), and their sum is (21), what will be the constant term?
Explanation opens after your attempt
Step 1
Concept
From (2r+5r=21), we get (r=3), so the roots are (6) and (15). The constant term is the product of roots (90).
Step 2
Why this answer is correct
The correct answer is A. (90). From (2r+5r=21), we get (r=3), so the roots are (6) and (15). The constant term is the product of roots (90).
Step 3
Exam Tip
(2r+5r=21) से (r=3) मिलता है, इसलिए मूल (6) और (15) हैं। स्थिर पद मूलों का गुणनफल (90) होगा।
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Question
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 28
यदि किसी मोनिक द्विघात समीकरण के मूल (r) और (3r) हैं तथा उनका योग (16) है, तो उस समीकरण का स्थिर पद क्या होगा?
If the roots of a monic quadratic equation are (r) and (3r), and their sum is (16), what will be the constant term?
Explanation opens after your attempt
Step 1
Concept
From (r+3r=16), we get (r=4), so the roots are (4) and (12). The constant term is the product of roots (48).
Step 2
Why this answer is correct
The correct answer is A. (48). From (r+3r=16), we get (r=4), so the roots are (4) and (12). The constant term is the product of roots (48).
Step 3
Exam Tip
(r+3r=16) से (r=4) मिलता है, इसलिए मूल (4) और (12) हैं। स्थिर पद मूलों का गुणनफल (48) होगा।
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Question
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि किसी मोनिक द्विघात समीकरण के मूल (r) और (2r) हैं तथा उनका योग (9) है, तो उस समीकरण में स्थिर पद क्या होगा?
If the roots of a monic quadratic equation are (r) and (2r), and their sum is (9), what will be the constant term?
Explanation opens after your attempt
Step 1
Concept
From (r+2r=9), we get (r=3), so the roots are (3) and (6). The constant term will be the product of roots (18).
Step 2
Why this answer is correct
The correct answer is A. (18). From (r+2r=9), we get (r=3), so the roots are (3) and (6). The constant term will be the product of roots (18).
Step 3
Exam Tip
(r+2r=9) से (r=3) मिलता है, इसलिए मूल (3) और (6) हैं। स्थिर पद मूलों का गुणनफल (18) होगा।
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