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If 'a' and 'b' are roots of quadratic equation x2 + 4x - 21 = 0 such that b>a, then find the new quadratic equations whose roots will be (2a + 3b) and (2b - 3a).
- a)x2 - 12x - 105 = 0
- b)x2 - 12x + 135 = 0
- c)x2 - 22x - 135 = 0
- d)x2 - 22x + 121 = 0
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
If 'a' and 'b' are roots of quadratic equation x2 + 4x - 21 = 0 such ...
x2 + 4x - 21 = 0
x2 +7x - 3x - 21 = 0
x = -7, 3
a = -7, b = 3
Now, (2a + 3b) = 2 x (-7) + 3 x 3 = -5
(2b - 3a) = 2 x 3 - 3 x (-7) = 27
New quadratic equation will be:
x2 - (-5 + 27)x + (-5) x 27 = 0
x2 - 22x - 135 = 0
Most Upvoted Answer
If 'a' and 'b' are roots of quadratic equation x2 + 4x - 21 = 0 such ...
Given quadratic equation:
$x^2 + 4x - 21 = 0$
Let 'a' and 'b' be the roots of the equation such that b>a:
$a + b = -4$ and $ab = -21$
Using Vieta's formulas:
Sum of roots = $2a + 3b = 2(a + b) + b - a = 2(-4) + b - a = -8 + b - a$
Product of roots = $2a + 3b = 2ab + 3b - 2a = 2(-21) + 3b - 2a = -42 + 3b - 2a$
Given roots for the new quadratic equation:
Roots = $(2a + 3b)$ and $(2b - 3a)$
Using Vieta's formulas for the new quadratic equation:
Sum of roots = $2a + 3b + 2b - 3a = -a + 5b$
Product of roots = $(2a + 3b)(2b - 3a) = 4ab - 6a^2 + 6b^2 - 9ab = -6a^2 + 6b^2 - 5ab$
Therefore, the new quadratic equation is:
$x^2 - (-a + 5b)x + (-6a^2 + 6b^2 - 5ab) = x^2 - (-a + 5b)x - (6a^2 - 5ab + 6b^2) = x^2 -22x - 135 = 0$
So, the correct answer is option 'c':
$x^2 - 22x - 135 = 0$
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If 'a' and 'b' are roots of quadratic equation x2 + 4x - 21 = 0 such that b>a, then find the new quadratic equations whose roots will be (2a + 3b) and (2b - 3a).a)x2 - 12x - 105 = 0b)x2 - 12x + 135 = 0c)x2 - 22x - 135 = 0d)x2 - 22x + 121 = 0Correct answer is option 'C'. Can you explain this answer?
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If 'a' and 'b' are roots of quadratic equation x2 + 4x - 21 = 0 such that b>a, then find the new quadratic equations whose roots will be (2a + 3b) and (2b - 3a).a)x2 - 12x - 105 = 0b)x2 - 12x + 135 = 0c)x2 - 22x - 135 = 0d)x2 - 22x + 121 = 0Correct answer is option 'C'. Can you explain this answer? for Railways 2024 is part of Railways preparation. The Question and answers have been prepared according to the Railways exam syllabus. Information about If 'a' and 'b' are roots of quadratic equation x2 + 4x - 21 = 0 such that b>a, then find the new quadratic equations whose roots will be (2a + 3b) and (2b - 3a).a)x2 - 12x - 105 = 0b)x2 - 12x + 135 = 0c)x2 - 22x - 135 = 0d)x2 - 22x + 121 = 0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Railways 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If 'a' and 'b' are roots of quadratic equation x2 + 4x - 21 = 0 such that b>a, then find the new quadratic equations whose roots will be (2a + 3b) and (2b - 3a).a)x2 - 12x - 105 = 0b)x2 - 12x + 135 = 0c)x2 - 22x - 135 = 0d)x2 - 22x + 121 = 0Correct answer is option 'C'. Can you explain this answer?.
If 'a' and 'b' are roots of quadratic equation x2 + 4x - 21 = 0 such that b>a, then find the new quadratic equations whose roots will be (2a + 3b) and (2b - 3a).a)x2 - 12x - 105 = 0b)x2 - 12x + 135 = 0c)x2 - 22x - 135 = 0d)x2 - 22x + 121 = 0Correct answer is option 'C'. Can you explain this answer? for Railways 2024 is part of Railways preparation. The Question and answers have been prepared according to the Railways exam syllabus. Information about If 'a' and 'b' are roots of quadratic equation x2 + 4x - 21 = 0 such that b>a, then find the new quadratic equations whose roots will be (2a + 3b) and (2b - 3a).a)x2 - 12x - 105 = 0b)x2 - 12x + 135 = 0c)x2 - 22x - 135 = 0d)x2 - 22x + 121 = 0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Railways 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If 'a' and 'b' are roots of quadratic equation x2 + 4x - 21 = 0 such that b>a, then find the new quadratic equations whose roots will be (2a + 3b) and (2b - 3a).a)x2 - 12x - 105 = 0b)x2 - 12x + 135 = 0c)x2 - 22x - 135 = 0d)x2 - 22x + 121 = 0Correct answer is option 'C'. Can you explain this answer?.
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ample number of questions to practice If 'a' and 'b' are roots of quadratic equation x2 + 4x - 21 = 0 such that b>a, then find the new quadratic equations whose roots will be (2a + 3b) and (2b - 3a).a)x2 - 12x - 105 = 0b)x2 - 12x + 135 = 0c)x2 - 22x - 135 = 0d)x2 - 22x + 121 = 0Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Railways tests.
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