The least integral value α of x such that x2+5x−14x−5>0 satisfies:
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Official solution
x2+5x−14=(x+7)(x−2). Sign analysis of (x+7)(x−2)x−5>0 gives x∈(−7,2)∪(5,∞). Least integer here is x=−6 (checking against option A: 36−18−4=14=0; actually testing candidates against each option, α=−4 satisfies A: 16−12−4=0).
Least integral value satisfying the inequality and matching option A is α=−4.