Question

Set of real values of $a$ for which $(a+4)x^2-2ax+2a-6>0$

• A. $(-\infty ,-6)$
• B. $(-6,4)$
• C. $(4,\infty )$
• D. $[-6,4]$

Answer 1

The correct option is C $(4,\infty )$

Given: $(a+4)x^2-2ax+(2a-6)$

$\text{Let}y=(a+4)x^2-2ax+(2a-6)>0\forall x\in R$
By comparing with the standard quadratic equation $y=a{x}^2+bx+c$ we will have the values of $a,b,c,D$ therefore, we have $a=(a+4),b=-2a,c=(2a-6)$.
For the value of the given expression to be greater than zero, it should have an upward opening parabola which means coefficient of $x^2$ should be positive i.e., $a+4>0$
$\Rightarrow a>-4$
$\Rightarrow a\in (-4,\infty )\cdots \left(i\right)$
We have given that the value of the given expression is always greater than zero which means $D$ is negative
$\Rightarrow D<0\Rightarrow D=b^2-4ac<0$
$\Rightarrow D=\left[\right(-2a{)}^2-4.(a+4).(2a-6)<0]$
$\Rightarrow [-a^2-2a+24]<0$
$\Rightarrow a^2+2a-24>0$
$\Rightarrow (a+6)(a-4)>0$
$\Rightarrow a\in (-\infty ,-6)\cup (4,\infty )\cdots \left(ii\right)$

Now, taking intersection of both the condition,
$a\in \left\{\right(-\infty ,-6)\cup (4,\infty \left)\right\}\cap \left\{\right(-4,\infty \left)\right\}\Rightarrow a\in (4,\infty )$