Question

For the given quadratic equation $y=-x^2-5x-4,y<0$ for

• A. $x\in (-4,-1)$
• B. $x\in (-\infty ,1)\cup (4,\infty )$
• C. $x\in (1,4)$
• D. $x\in (-\infty ,-4)\cup (-1,\infty )$

Answer 1

The correct option is D $x\in (-\infty ,-4)\cup (-1,\infty )$

Given: $y=-x^2-5x-4$
On comparing with the standard quadratic expression $y=a{x}^2+bx+c,$ we get $a=-1,b=-5,c=-4$.
And we know, $D=b^2-4ac=(-5{)}^2-4(-1\left)\right(-4)$
$D=9>0$
$\Rightarrow$ the equation will have two distinct roots.
Which will be:
$x=\frac{-(-5)\pm \sqrt{9}}{-2}\Rightarrow x=\frac{5+3}{-2}\text{or}x=\frac{5-3}{-2}\Rightarrow x=-4\text{or}x=-1$

Now $a<0\Rightarrow$ parabola will be downward opening; and $D>0\Rightarrow$ parabola will cut the $x$-axis at two distinct points.
$\therefore$ the graph of the given expression can be drawn as:


And given expression is negative for $x=(-\infty ,-4)\cup (-1,\infty )$