How do you solve quadratic inequalities/functions?

Question

Find the least value of $n$ such that $(n-2)x^2+8x+n+4 > 0$, where $x \in \mathbb R$ and $n \in \mathbb N$.

What I did -

As $x$ is real, $D>0$

$64-4(n+4)(n-2)>0$

$(n-4)(n-6)<0$

$n \epsilon (-6,4)$

But as $n \epsilon N$

The least value is 3

According to the book's method,

$n-2>0$

$n>2$

And,

$64-4(n-2)(n+4)<0$

$(n-4)(n+6)>0$

$n>4$

And least value of n comes out to be 5.

I don't understand why D<0 when x is real?

Thanks.

Answer 1

the case $n=0$ gives $$x^2-4x-2<0$$ which is not fulfilled for all real $x$ $$n=1$$ gives $$x^2-8x-5<0$$ $$n=2$$ gives $$8x+6>0$$ for $$n>2$$ we get $$x^2+2\left(\frac{4}{n-2}\right)x+\left(\frac{4}{n-2}\right)^2+\frac{n+4}{n-2}-\left(\frac{4}{n-2}\right)^2>0$$ and we get $$\left(x+\frac{4}{n-2}\right)^2+\frac{n^2+2n-24}{(n-2)^2}>0$$ finally we get the answer $$n=5$$ then we have $$\left(x+\frac{4}{3}\right)^2+\frac{11}{9}>0$$

Answer 2

Let's define $D=64−4(n+4)(n−2)$ and $f(x)=(n−2)x^2+8x+n+4$. If $D>0$ then $f(x)$ vanishes for two values of $x$. But we don't want $f(x)=0$, we need instead $f(x)>0$ for every $x$. That can happen only if $D<0$.