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Theory of Equations: Number Integral Values Graph 16x Always Above Axis

JEE Maths question with a full step-by-step solution.

Question
The number of integral values of aa so that the graph of y=16x2+8(a+5)x−7a−5y = 16x^2 + 8(a+5)x - 7a - 5 is always above the xx-axis is
A1212correct
B1313
C1414
D1515
Solution
Step 1: For the parabola to lie entirely above the xx-axis, the discriminant must be negative (note leading coefficient 16>016 > 0):
D=[8(a+5)]2−4⋅16⋅(−7a−5)<0D = \bigl[8(a+5)\bigr]^2 - 4 \cdot 16 \cdot (-7a-5) < 0
64(a+5)2+64(7a+5)<0  ⟹  (a+5)2+7a+5<064(a+5)^2 + 64(7a+5) < 0 \implies (a+5)^2 + 7a + 5 < 0
a2+17a+30<0  ⟹  (a+15)(a+2)<0  ⟹  −15<a<−2a^2 + 17a + 30 < 0 \implies (a+15)(a+2) < 0 \implies -15 < a < -2
 Step 2: Integer values: a∈{−14,−13,…,−3}a \in \{-14, -13, \ldots, -3\}, giving 1212 values. Correct answer: (1)
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