The number of integral value(s) of a so that the graph of y=16x2+8(a+5)x−7a−5 is always above the x-axis is

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answer is 12.

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Detailed Solution

We have 16x2+8(a+5)x−(7a+5)>0,∀x∈RSo, discriminant must be negative. Now, Discriminant =D=64(a+5)2+7a+5 As D<0⇒a2+17a+30<0⇒(a+15)(a+2)<0∴−15

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