Question

The least integral value \(\alpha\) of \(x\) such that \(\dfrac{x-5}{x^2+5x-14} \gt 0\), satisfies :

• A. \(\alpha^2 + 3\alpha - 4 = 0\)
• B. \(\alpha^2 - 5\alpha + 4 = 0\)
• C. \(\alpha^2 - 7\alpha + 6 = 0\)
• D. \(\alpha^2 + 5\alpha - 6 = 0\)

Hint:

Factorise the denominator, use the sign scheme (wavy curve), find the solution set and its least integer.

Answer 1

The Correct Option is D

Test the integers directly.
For \(x=-7\) the denominator is 0 (undefined); for \(x \le -8\), e.g. \(x=-8\): \(\frac{-13}{(-1)(-10)} = -1.3 \lt 0\), rejected.
For \(x=-6\): numerator \(-11\), denominator \((1)(-8) = -8\), ratio \(= 1.375 \gt 0\), accepted. Hence \(-6\) is the least integer.
Substituting \(\alpha=-6\) in the options, only \(\alpha^2+5\alpha-6 = 36-30-6 = 0\) holds.
Answer: (4) \(\alpha^2 + 5\alpha - 6 = 0\)