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- Subject
- Quadratic Equationsmathematics-mcqs › quadratic-equations
- Published
- 11 Jul 2019
- Last updated
- 28 May 2026
Explanation
For equation I: x² + 11x + 30 = 0 can be factored as (x + 6)(x + 5) = 0, giving solutions x = -6 and x = -5. For equation II: y² + 15y + 56 = 0 factors to (y + 8)(y + 7) = 0, so y = -8 and y = -7. Comparing the roots, the smallest x (-6) is greater than the largest y (-7), so x > y.
More Quadratic Equations MCQs
Practice related questions from the same subject.
- 1.What are the solutions to the quadratic equation 3x² - 7x - 6 = 0?
- 2.What are the solutions to the quadratic equation x² + x - 42 = 0?
- 3.What are the solutions of the quadratic equation 2x² + 5x + 2 = 0?
- 4.Given the quadratic equations I. a² - 13a + 42 = 0 and II. b² - 15b + 56 = 0, determine the correct relationship between the values of a and b obtained by solving these equations.
- 5.Given the equations: (i) a³ − 988 = 343 and (ii) b² − 72 = 49, determine the correct relationship between the values of a and b.
- 6.Given the quadratic equations I. 9a² + 18a + 5 = 0 and II. 2b² + 13b + 20 = 0, determine the correct relationship between the values of a and b after solving both equations.
- 7.Given the quadratic equations I. x² + 3x - 18 = 0 and II. y² + y - 30 = 0, what can be concluded about the relationship between x and y after solving both?
- 8.Given the equations I. x² + 9x + 20 = 0 and II. y² + 5y + 6 = 0, determine the relationship between the values of x and y after solving both equations.
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