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The sum of the two real roots of the equation x & # 178; + 2 (K + 1) x + K & # 178; = 0 of X is m, and M = - 2 (K + 1), If the inequality system {y ＞ - 4, y ＜ m} of Y has a real solution, then the value range of K is ()
Hello!
x² + 2(k+1)x + k² = 0
Δ = 4(k+1)² - 4k² ≥ 0
k ≥ - 1/2
System of inequalities - 4 < y < m has solution
Then M > - 4
That is - 2 (K + 1) > - 4
k < 1
∴ - 1/2 ≤ k < 1
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If the equation x & # 178; + (m-1) x + 2 = 0 of X has two unequal real roots, then the value of K is?
(m-1)²-4>0
m²-2m-3>0
(m-3)(m+1)>0
Ψ m > 3 or M
If the sum of the two real roots of the equation x2 + 2 (K + 1) x + K2 = 0 is m, and M = - 2 (K + 1), the inequality system y ＞ - 4Y < m of Y has a real solution, then the value range of K is___ ．
∵ the equation x2 + 2 (K + 1) x + K2 = 0 has two real roots, and the solution is k ≥ - 12; the inequality system y ＞ - 4Y ＜ m about y has a real number solution, and the solution is k ＜ 1. The range of K is - 12 ≤ K ＜ 1. So fill in the blanks: - 12 ≤ K ＜ 1
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