The inequality that represents the range of k so that inequality 1 has real solution is -3 < k < 5
Inequality 1
x^2 + (k + 1)x + k + 4 > 0
The discriminant is calculated using
D = b^2 - 4ac
So, we have:
D = (k + 1)^2 - 4 * 1 * (k + 4)
Evaluate
D = k^2 + 2k + 1 - 4k - 16
This gives
D = k^2- 2k - 15
Factorize the equation
D = (k+ 3)(k -5)
Solve for k
k = -3 and k = 5
Express the k values as inequalities
-3 < k < 5
Inequality 2
The inequality cannot be read, because of the image quality
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