Question

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

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

Answer 1

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

Let $y=(a+4)x^2-2ax+(2a-6)>0\forall x\in R$
By comparing with the standard quadratic equation $\text{y}=a{\text{x}}^2+b\text{x}+c$ 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., $a+4>0$
$\Rightarrow a>-4$
$\Rightarrow a\in (-4,\infty )\cdot \cdot \cdot \left(i\right)$
We have given that the value of the given expression is always greater than zero which means $D$ is negative
$\Rightarrow D<0\Rightarrow D=b^2-4ac<0$
$\Rightarrow D=\left[\right(-2a{)}^2-4.(a+4).(2a-6)<0]$
$\Rightarrow [-a^2-2a+24]<0$
$\Rightarrow a^2+2a-24>0$
$\Rightarrow (a+6)(a-4)>0$
$\Rightarrow a\in (-\infty ,-6)\cup (4,\infty )\cdot \cdot \cdot \left(ii\right)$

From (i) and (ii) we will get $a\in (4,\infty )$