CA Foundation Exam > CA Foundation Questions > Os. 16. If p? =5p-3 and q? =5q-3, where p is ...
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Os. 16. If p? =5p-3 and q? =5q-3, where p is not equals to a, then the quadratic equation whose roots are p/q and a/pis a. ×2-19× 3=0 b. 3x2-19x-3=0 c. 3x7-19x 3=0 d. 3x? 19× 3=0?
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Os. 16. If p? =5p-3 and q? =5q-3, where p is not equals to a, then the...
To find the quadratic equation whose roots are p/q and a/p, we can use the Vieta's formulas.
Let's first find the sum and product of the roots:
- The sum of the roots, p/q and a/p, is given by:
(p/q) + (a/p) = (p^2 + aq) / (pq)
- The product of the roots, p/q and a/p, is given by:
(p/q) * (a/p) = (ap) / (pq)
According to Vieta's formulas, these values are related to the coefficients of the quadratic equation.
The quadratic equation can be written in the form:
x^2 - (sum of roots)x + (product of roots) = 0
So, we need to find the sum and product of the roots and substitute them into the equation.
Let's calculate the sum and product of the roots using the given expressions for p? and q?:
- The sum of the roots, p/q and a/p, is:
(p^2 + aq) / (pq) = [(5p-3)^2 + a(5q-3)] / [(5p-3)(5q-3)]
- The product of the roots, p/q and a/p, is:
(ap) / (pq) = (a(5p-3)) / [(5p-3)(5q-3)]
Now, we can substitute these values into the quadratic equation:
x^2 - [(p^2 + aq) / (pq)]x + [(ap) / (pq)] = 0
Simplifying the equation further:
x^2 - (p^2 + aq)x / (pq) + (ap)x / (pq) = 0
Multiplying through by (pq) to eliminate the denominators:
(pq)x^2 - (p^2 + aq)x + (ap)x = 0
Finally, rearranging the terms, we get the quadratic equation:
pqx^2 - (p^2 + aq)x + (ap)x = 0
The correct option is b. 3x^2 - 19x - 3 = 0.
Let's first find the sum and product of the roots:
- The sum of the roots, p/q and a/p, is given by:
(p/q) + (a/p) = (p^2 + aq) / (pq)
- The product of the roots, p/q and a/p, is given by:
(p/q) * (a/p) = (ap) / (pq)
According to Vieta's formulas, these values are related to the coefficients of the quadratic equation.
The quadratic equation can be written in the form:
x^2 - (sum of roots)x + (product of roots) = 0
So, we need to find the sum and product of the roots and substitute them into the equation.
Let's calculate the sum and product of the roots using the given expressions for p? and q?:
- The sum of the roots, p/q and a/p, is:
(p^2 + aq) / (pq) = [(5p-3)^2 + a(5q-3)] / [(5p-3)(5q-3)]
- The product of the roots, p/q and a/p, is:
(ap) / (pq) = (a(5p-3)) / [(5p-3)(5q-3)]
Now, we can substitute these values into the quadratic equation:
x^2 - [(p^2 + aq) / (pq)]x + [(ap) / (pq)] = 0
Simplifying the equation further:
x^2 - (p^2 + aq)x / (pq) + (ap)x / (pq) = 0
Multiplying through by (pq) to eliminate the denominators:
(pq)x^2 - (p^2 + aq)x + (ap)x = 0
Finally, rearranging the terms, we get the quadratic equation:
pqx^2 - (p^2 + aq)x + (ap)x = 0
The correct option is b. 3x^2 - 19x - 3 = 0.
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Os. 16. If p? =5p-3 and q? =5q-3, where p is not equals to a, then the quadratic equation whose roots are p/q and a/pis a. ×2-19× 3=0 b. 3x2-19x-3=0 c. 3x7-19x 3=0 d. 3x? 19× 3=0?
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Os. 16. If p? =5p-3 and q? =5q-3, where p is not equals to a, then the quadratic equation whose roots are p/q and a/pis a. ×2-19× 3=0 b. 3x2-19x-3=0 c. 3x7-19x 3=0 d. 3x? 19× 3=0? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about Os. 16. If p? =5p-3 and q? =5q-3, where p is not equals to a, then the quadratic equation whose roots are p/q and a/pis a. ×2-19× 3=0 b. 3x2-19x-3=0 c. 3x7-19x 3=0 d. 3x? 19× 3=0? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Os. 16. If p? =5p-3 and q? =5q-3, where p is not equals to a, then the quadratic equation whose roots are p/q and a/pis a. ×2-19× 3=0 b. 3x2-19x-3=0 c. 3x7-19x 3=0 d. 3x? 19× 3=0?.
Os. 16. If p? =5p-3 and q? =5q-3, where p is not equals to a, then the quadratic equation whose roots are p/q and a/pis a. ×2-19× 3=0 b. 3x2-19x-3=0 c. 3x7-19x 3=0 d. 3x? 19× 3=0? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about Os. 16. If p? =5p-3 and q? =5q-3, where p is not equals to a, then the quadratic equation whose roots are p/q and a/pis a. ×2-19× 3=0 b. 3x2-19x-3=0 c. 3x7-19x 3=0 d. 3x? 19× 3=0? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Os. 16. If p? =5p-3 and q? =5q-3, where p is not equals to a, then the quadratic equation whose roots are p/q and a/pis a. ×2-19× 3=0 b. 3x2-19x-3=0 c. 3x7-19x 3=0 d. 3x? 19× 3=0?.
Solutions for Os. 16. If p? =5p-3 and q? =5q-3, where p is not equals to a, then the quadratic equation whose roots are p/q and a/pis a. ×2-19× 3=0 b. 3x2-19x-3=0 c. 3x7-19x 3=0 d. 3x? 19× 3=0? in English & in Hindi are available as part of our courses for CA Foundation.
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Os. 16. If p? =5p-3 and q? =5q-3, where p is not equals to a, then the quadratic equation whose roots are p/q and a/pis a. ×2-19× 3=0 b. 3x2-19x-3=0 c. 3x7-19x 3=0 d. 3x? 19× 3=0?, a detailed solution for Os. 16. If p? =5p-3 and q? =5q-3, where p is not equals to a, then the quadratic equation whose roots are p/q and a/pis a. ×2-19× 3=0 b. 3x2-19x-3=0 c. 3x7-19x 3=0 d. 3x? 19× 3=0? has been provided alongside types of Os. 16. If p? =5p-3 and q? =5q-3, where p is not equals to a, then the quadratic equation whose roots are p/q and a/pis a. ×2-19× 3=0 b. 3x2-19x-3=0 c. 3x7-19x 3=0 d. 3x? 19× 3=0? theory, EduRev gives you an
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