Question

For what values of $x$ is $$\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0?$$ note: be thorough and explain why all points in your answer are solutions and why all points outside your answer are not solutions.

Answer 1

To solve the inequality $\frac{x^2+x+3}{2x^2+x-6}\ge \:0$

$\mathrm{Factor\:the\:left\:hand\:side\:}\frac{x^2+x+3}{2x^2+x-6}:$

The numerator $x^2+x+3$ is not factorizable.

so factor the denominator $2x^2+x-6$:

$2x^2+x-6=\left(2x^2-3x\right)+\left(4x-6\right)=x\left(2x-3\right)+2\left(2x-3\right)\\ \mathrm{Factor\:out\:}x\mathrm{\:from\:}2x^2-3x\mathrm{:\quad }x\left(2x-3\right)\\ \mathrm{Factor\:out\:}2\mathrm{\:from\:}4x-6\mathrm{:\quad }2\left(2x-3\right)$

Now take $\mathrm{Factor\:out\:common\:term\:}\left(2x-3\right)$

Then we get factor of the denominator as $\left(2x-3\right)\left(x+2\right)$

Thus $\frac{x^2+x+3}{2x^2+x-6}=\frac{x^2+x+3}{\left(x+2\right)\left(2x-3\right)}$

Now $\mathrm{Compute\:the\:signs\:of\:the\:factors\:of\:}\frac{x^2+x+3}{\left(x+2\right)\left(2x-3\right)}$

Signs of $x^2+x+3>0$

$\mathrm{Choosing\:ranges\:that\:satisfy\:the\:required\:condition:}\:\ge \:\:0$

$x\frac{3}{2}$ is the required solution of the given inequality.

Answer 2

Answer: $(-\infty, -2)\cup(3/2,\infty)$

Step-by-step explanation:

Here, the given expression,

$\frac{x^2 + x + 3}{2x^2 + x - 6} \geq 0$

Since, for the value of x,

Denominator ≠ 0,

$2x^2+x-6\neq 0$

$\implies 2x^2+4x - 3x - 6\neq =0$

$\implies 2x(x+2)-3(x+2)\neq 0$

$\implies (2x-3)(x+2)\neq 0$

$\text{ if }2x-3\neq 0\implies x\neq \frac{3}{2}$

$\text{If } x+2\neq 0\implies x\neq -2$

Thus, there is three intervals possible,

1) $(-\infty,-2)$

2) $(-2,\frac{3}{2})$

3) $(\frac{3}{2},\infty)$

In first intervals, $\frac{x^2 + x + 3}{2x^2 + x - 6} \geq 0$ is  true.

⇒ $(-\infty,-2)$ will contain the value of x.

In second interval, $\frac{x^2 + x + 3}{2x^2 + x - 6} \geq 0$

is not true,

⇒ $(-2, -\frac{3}{2})$ will not contain the value of x,

In third interval,

In third interval, $\frac{x^2 + x + 3}{2x^2 + x - 6} \geq 0$

is true,

⇒ $(-2, -\frac{3}{2})$ will contain the value of x,

In third interval,

Thus, the value of x is,

$(-\infty, -2)\cup(3/2,\infty)$