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If alpha , beta are the roots of the equation x^2 -5x 6=0 and alpha>beta then the equation with the roots (alpha×beta alpha beta) and (alpha×beta-alpha-beta)?
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If alpha , beta are the roots of the equation x^2 -5x 6=0 and alpha>be...
Solution:
Given equation is x^2 - 5x + 6 = 0
We can factorize the above equation as follows:
(x - 2)(x - 3) = 0
Therefore, the roots of the equation are x = 2 and x = 3.
We are given that alpha > beta. Therefore, alpha = 3 and beta = 2.
Now, we need to find the equation with roots (alpha×beta alpha beta) and (alpha×beta-alpha-beta).
Let us first find the values of the roots:
(alpha×beta alpha beta) = (3×2 3 2) = (6 3 2)
(alpha×beta-alpha-beta) = (3×2-3-2) = (3 -5)
Therefore, the roots of the required equation are 6, 3, 2, 3, and -5.
Let us now find the equation with these roots:
We know that a quadratic equation with roots α and β can be written as:
(x - α)(x - β) = 0
Similarly, a quadratic equation with roots αβ, α, β can be written as:
(x - αβ)(x - α)(x - β) = 0
Substituting the values of α, β, and αβ, we get:
(x - 6)(x - 3)(x - 2) = 0
Expanding the above equation, we get:
x^3 - 11x^2 + 36x - 36 = 0
Therefore, the required equation is x^3 - 11x^2 + 36x - 36 = 0.
Hence, the answer is x^3 - 11x^2 + 36x - 36 = 0.
Given equation is x^2 - 5x + 6 = 0
We can factorize the above equation as follows:
(x - 2)(x - 3) = 0
Therefore, the roots of the equation are x = 2 and x = 3.
We are given that alpha > beta. Therefore, alpha = 3 and beta = 2.
Now, we need to find the equation with roots (alpha×beta alpha beta) and (alpha×beta-alpha-beta).
Let us first find the values of the roots:
(alpha×beta alpha beta) = (3×2 3 2) = (6 3 2)
(alpha×beta-alpha-beta) = (3×2-3-2) = (3 -5)
Therefore, the roots of the required equation are 6, 3, 2, 3, and -5.
Let us now find the equation with these roots:
We know that a quadratic equation with roots α and β can be written as:
(x - α)(x - β) = 0
Similarly, a quadratic equation with roots αβ, α, β can be written as:
(x - αβ)(x - α)(x - β) = 0
Substituting the values of α, β, and αβ, we get:
(x - 6)(x - 3)(x - 2) = 0
Expanding the above equation, we get:
x^3 - 11x^2 + 36x - 36 = 0
Therefore, the required equation is x^3 - 11x^2 + 36x - 36 = 0.
Hence, the answer is x^3 - 11x^2 + 36x - 36 = 0.
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If alpha , beta are the roots of the equation x^2 -5x 6=0 and alpha>beta then the equation with the roots (alpha×beta alpha beta) and (alpha×beta-alpha-beta)?
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If alpha , beta are the roots of the equation x^2 -5x 6=0 and alpha>beta then the equation with the roots (alpha×beta alpha beta) and (alpha×beta-alpha-beta)? 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 If alpha , beta are the roots of the equation x^2 -5x 6=0 and alpha>beta then the equation with the roots (alpha×beta alpha beta) and (alpha×beta-alpha-beta)? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If alpha , beta are the roots of the equation x^2 -5x 6=0 and alpha>beta then the equation with the roots (alpha×beta alpha beta) and (alpha×beta-alpha-beta)?.
If alpha , beta are the roots of the equation x^2 -5x 6=0 and alpha>beta then the equation with the roots (alpha×beta alpha beta) and (alpha×beta-alpha-beta)? 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 If alpha , beta are the roots of the equation x^2 -5x 6=0 and alpha>beta then the equation with the roots (alpha×beta alpha beta) and (alpha×beta-alpha-beta)? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If alpha , beta are the roots of the equation x^2 -5x 6=0 and alpha>beta then the equation with the roots (alpha×beta alpha beta) and (alpha×beta-alpha-beta)?.
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