141.삼차방정식 $x^{3}+7 x^{2}+(6-a) x-a=0$ 의 서로 다른 실근의 개수가 2 개가 되도록 하는 모든 실수 $a$ 의 값의 합은?[4.6점]
(1) -14
(2) -13
(3) -12
(4) -11
(5) -10

Answer

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Answer

Calculate the sum of the real solutions: $\boxed{0.37}$

Steps

Step 1 ：Given the cubic equation: $x^3 + 7x^2 + (6-a)x - a = 0$

Step 2 ：Calculate the discriminant: $Δ = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2$ with $a = 1, b = 7, c = (6 - a), d = -a$

Step 3 ：Find the value of 'a' for which the discriminant is equal to zero

Step 4 ：Solve the equation: $-27a^4 - 126a^2(6 - a) - 4a(6 - a)^3 + 1372a + 49(6 - a)^2 = 0$

Step 5 ：Find the real solutions for 'a': $a \approx 2.383$ and $a \approx -2.010$

Step 6 ：Calculate the sum of the real solutions: $\boxed{0.37}$