Question

For all '$x$' , $x^2+2ax+(10-3a)>0,$ then the interval in which '$a$' lies is

• A. a < - 5
• B. - 5 < a < 2
• C. a > 5
• D. 2 < a < 5

Answer 1

Answer: (B)

As we know, $a{x}^2+bx+c>0$ for all $x\in R,$ iff $a>0$ and $D<0$
Given equation is $x^2+2ax+(10-3a)>0,\forall x\in R$
Now, $D<0$
$\Rightarrow 4a^2-4(10-3a)<0$
$\Rightarrow 4(a^2+3a-10)<0$
$\Rightarrow (a+5)(a+2)<0$
$\Rightarrow a\in (-5,2)$