Second Degree Equations — Important Questions
Kerala Board · Class 10 · Mathematics
30 important questions from Second Degree Equations for Kerala Board Class 10 Mathematics, with answers. Written for the board exams 2027.
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Important Questions from Second Degree Equations
Solve the equation (x - 5)² = 36. What are the possible values of x?
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x = 11, x = -1
From (x - 5)² = 36, we get x - 5 = ±6. So x - 5 = 6 gives x = 11, and x - 5 = -6 gives x = -1. Both solutions are valid algebraically.
Which of the following expressions can be written as perfect squares?
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x² + 6x + 9, x² - 10x + 25, x² - 8x + 16
x² + 6x + 9 = (x+3)², x² - 10x + 25 = (x-5)², x² - 8x + 16 = (x-4)². But x² + 4x + 5 cannot be written as a perfect square since 4² ≠ 4×4 = 16 ≠ 5×4 = 20.
A right triangle has perpendicular sides where one is 3 cm more than the other. If the area is 35 cm², what are the lengths of these sides?
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7 cm and 10 cm
Let the shorter side be x cm, longer side = (x+3) cm. Area = ½x(x+3) = 35, so x(x+3) = 70. This gives x² + 3x = 70. Completing square: x² + 3x + 9/4 = 70 + 9/4 = 289/4. So (x + 3/2)² = 289/4, x + 3/2 = 17/2, x = 7. Sides are 7 cm and 10 cm.
If x² - 12x + 36 = 0, what is the value of x?
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6
Notice that x² - 12x + 36 = (x-6)². So the equation becomes (x-6)² = 0, which means x-6 = 0, therefore x = 6. This is a perfect square trinomial with only one solution.
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Sources & Official References
- Kerala Board of Public Examinations — keralapareekshabhavan.in
- National Education Policy 2020 — education.gov.in
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