Question: Find the domain.
y = √(6x² + 11x - 7) / √(|2x - 5| + 4x² - 2x - 13)
graph_shape: none; position_hint: centerUser: find to domain and range
1. **State the problem.**
We need the domain of $$y=\frac{\sqrt{6x^2+11x-7}}{\sqrt{|2x-5|+4x^2-2x-13}}$$.
Because there are square roots in both the numerator and denominator, we must make sure:
- the expression inside the numerator square root is at least $0$,
- the expression inside the denominator square root is greater than $0$,
- and the denominator is not $0$.
2. **Set up the domain conditions.**
For the numerator:
$$6x^2+11x-7\ge 0$$
For the denominator:
$$|2x-5|+4x^2-2x-13>0$$
3. **Solve the numerator inequality.**
Factor the quadratic:
$$6x^2+11x-7=(6x-3)(x+\tfrac{7}{3})$$
A cleaner factorization is:
$$6x^2+11x-7=(3x+7)(2x-1)$$
So we solve:
$$(3x+7)(2x-1)\ge 0$$
The critical points are:
$$x=-\tfrac{7}{3},\; x=\tfrac{1}{2}$$
Since the parabola opens upward, the product is nonnegative outside the roots:
$$x\le -\tfrac{7}{3}\quad \text{or}\quad x\ge \tfrac{1}{2}$$
4. **Simplify the denominator expression by cases.**
We need:
$$|2x-5|+4x^2-2x-13>0$$
Use the absolute value split at $x=\tfrac{5}{2}$.
**Case 1: $x\ge \tfrac{5}{2}$**
Then $|2x-5|=2x-5$, so the denominator becomes:
$$2x-5+4x^2-2x-13=4x^2-18$$
So we need:
$$4x^2-18>0$$
$$4x^2>18$$
$$x^2>\tfrac{9}{2}$$
Since this case already has $x\ge \tfrac{5}{2}$, the condition becomes:
$$x>\tfrac{3}{\sqrt{2}}$$
**Case 2: $x<\tfrac{5}{2}$**
Then $|2x-5|=5-2x$, so the denominator becomes:
$$5-2x+4x^2-2x-13=4x^2-4x-8$$
Now solve:
$$4x^2-4x-8>0$$
Divide by $4$:
$$\frac{\cancel{4}x^2-\cancel{4}x-\cancel{8}}{\cancel{4}}>0$$
$$x^2-x-2>0$$
Factor:
$$(x-2)(x+1)>0$$
This is positive when:
$$x<-1\quad \text{or}\quad x>2$$
Because we are in the case $x<\tfrac{5}{2}$, both parts are allowed here.
So the denominator is positive for:
$$x<-1\quad \text{or}\quad 2\tfrac{3}{\sqrt{2}}$$
But notice $x>\tfrac{3}{\sqrt{2}}$ is already contained in $x>2$, so the combined denominator condition is simply:
$$x<-1\quad \text{or}\quad x>2$$
5. **Intersect the numerator and denominator conditions.**
Numerator condition:
$$x\le -\tfrac{7}{3}\quad \text{or}\quad x\ge \tfrac{1}{2}$$
Denominator condition:
$$x<-1\quad \text{or}\quad x>2$$
Now take the overlap:
- For the left side, $x\le -\tfrac{7}{3}$ already fits inside $x<-1$.
- For the right side, $x\ge \tfrac{1}{2}$ and $x>2$ together give $x>2$.
So the domain is:
$$(-\infty,-\tfrac{7}{3}]\cup(2,\infty)$$
6. **Final answer.**
The domain is $$(-\infty,-\tfrac{7}{3}]\cup(2,\infty)$$.