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# prove that no real value of k exists for the equation x^2-kx+2k-10=0 has equal roots.(has onlyone value for x)?

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- HemantLv 79 years agoFavorite Answer
QE : 1x² + (-k)x + (2k-10) = 0

∴ a = 1, b = -k, c = 2k - 10 .......... (1)

For equal roots, discriminant Δ = b² - 4ac = 0.

∴ from (1),

... (-k)² - 4(1)( 2k - 10 ) = 0

∴ 1k² - 8k + 20 = 0.

For this k-equation, discr. = (-8)² - 4(1)(20) = -14 < 0

∴ there is NO REAL value of k. ........................................ Q.E.D.

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