Question

If $x^2+2ax+10-3a>0$ for every real value of $x$, then

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

Answer 1

The correct option is C $-5<a<2$

Let $y=x^2+2ax+10-3a$

$=x^2+2ax+10-3a-a^2+a^2$

$=(x+a{)}^2-(a^2+3a-10)$

$=(x+a{)}^2-(a+5\left)\right(a-2)$

If $y>0$ for all $x$ then we need $(a+5)(a-2)<0[\because -(a+5\left)\right(a-2)>0\Rightarrow (x+a{)}^2-(a+5)(a-2)>0forallx]$

So, $(a+5)(a-2)<0$

$\Rightarrow -5<a<2$

So, $\text{C}$ is the correct option.