Given that the two real roots of equation x2 + KX + 6 = 0 are x1, X2, and the two real roots of equation x2-kx + 6 = 0 are X1 + 5, X2 + 5, then the value of K is equal to () A. 5B. -5C. 7D. -7

∵ the two real roots of equation x2 + KX + 6 = 0 are x1, X2, ∵ X1 + x2 = - K, x1 · x2 = 6; and the two real roots of equation x2-kx + 6 = 0 are X1 + 5, X2 + 5, ∵ X1 + 5 + x2 + 5 = k, (x1 + 5) · (x2 + 5) = 6, ∵ X1 + x2 = K-10, x1x2 + 5 (x1 + x2) + 25 = 6, ∵ K-10 = - K, the solution is k = 5

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