Asked in: July 2024 board paper; the same “a root is given, find the constant” type came in March 2024 (3 is a root of kx² - 7x + 12 = 0 ), March 2026 ( -7/5 is a root of 5m² + 2m + k = 0) and June 2026 (2 is a root of x² + mx - 5 = 0) Find ‘ k ’ if x = 3 is a root of quadratic equation kx² - 10x + 3 = 0.

Chapter: Quadratic Equations· Mathematics Part 1 (Algebra)

Asked in: July 2024 board paper; the same “a root is given, find the constant” type came in March 2024 (3 is a root of \(kx^2 - 7x + 12 = 0\)), March 2026 (\(-\dfrac{7}{5}\) is a root of \(5m^2 + 2m + k = 0\)) and June 2026 (2 is a root of \(x^2 + mx…

Asked in: July 2024 board paper; the same “a root is given, find the constant” type came in March 2024 (3 is a root of \(kx^2 - 7x + 12 = 0\)), March 2026 (\(-\dfrac{7}{5}\) is a root of \(5m^2 + 2m + k = 0\)) and June 2026 (2 is a root of \(x^2 + mx - 5 = 0\))

Find ‘\(k\)’ if \(x = 3\) is a root of quadratic equation \(kx^2 - 10x + 3 = 0\).

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Answer: \(k = 3\)

A root satisfies the equation. Put \(x = 3\):

\(k(3)^2 - 10(3) + 3 = 0\)

\(\therefore 9k - 30 + 3 = 0\)

\(\therefore 9k = 27\)

\(\therefore k = 3\)

(The same method gives: March 2024, \(9k - 21 + 12 = 0 \Rightarrow k = 1\); March 2026, \(5 \times \dfrac{49}{25} - \dfrac{14}{5} + k = 0 \Rightarrow 7 + k = 0 \Rightarrow k = -7\); June 2026, \(4 + 2m - 5 = 0 \Rightarrow m = \dfrac{1}{2}\).)

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Asked in: July 2024 board paper; the same “a root is given, find the constant” type came in March 2024 (3 is a root of \(kx^2 - 7x + 12 = 0\)), March 2026 (\(-\dfrac{7}{5}\) is a root of \(5m^2 + 2m + k = 0\)) and June 2026 (2 is a root of \(x^2 + mx…

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