Question

The graph of the function $y=16x^2+8(a+5)x-7a-5$ is strictly above the X-axis, then $a$ must satisfy the inequality

• A. $-15<a<-2$
• B. $-2<a<-1$
• C. $5<a<7$
• D. $2<a<15$

Answer 1

Answer: (A)

$-15<a<-2$

$b^2-4ac<0$
$64(a+5{)}^2+64(7a+5)<0$
$a^2+25+10a+7a+5<0$
$a^2+17a+30<0$
$(a+15)(a+2)<0$
So, the value of $a$ for which the function will always be above X-axis is $-15<a<-2$ .