Question: Find the domain.
y = \frac{\sqrt{6x^2 + 11x - 7}}{\sqrt{|2x - 5| + 4x^2 - 2x - 13}}
1. **State the problem:** Find the domain of the function $$y = \frac{\sqrt{6x^2 + 11x - 7}}{\sqrt{|2x - 5| + 4x^2 - 2x - 13}}$$.
2. **Recall domain rules for such functions:**
- The expression inside any square root must be \(\geq 0\) because the square root of a negative number is not real.
- The denominator cannot be zero because division by zero is undefined.
3. **Set conditions for the numerator:**
$$6x^2 + 11x - 7 \geq 0$$
4. **Solve the quadratic inequality for the numerator:**
Find roots of $$6x^2 + 11x - 7 = 0$$ using the quadratic formula:
$$x = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 6 \cdot (-7)}}{2 \cdot 6} = \frac{-11 \pm \sqrt{121 + 168}}{12} = \frac{-11 \pm \sqrt{289}}{12} = \frac{-11 \pm 17}{12}$$
Roots:
$$x_1 = \frac{-11 - 17}{12} = \frac{-28}{12} = -\frac{7}{3}$$
$$x_2 = \frac{-11 + 17}{12} = \frac{6}{12} = \frac{1}{2}$$
Since the leading coefficient \(6 > 0\), the parabola opens upward, so the quadratic is \(\geq 0\) outside the roots:
$$x \leq -\frac{7}{3} \quad \text{or} \quad x \geq \frac{1}{2}$$
5. **Set conditions for the denominator:**
$$\sqrt{|2x - 5| + 4x^2 - 2x - 13} \neq 0$$
So,
$$|2x - 5| + 4x^2 - 2x - 13 > 0$$
6. **Analyze the expression inside the denominator's square root:**
Let
$$D = |2x - 5| + 4x^2 - 2x - 13$$
We want $$D > 0$$.
7. **Consider cases for the absolute value:**
- Case 1: $$2x - 5 \geq 0 \Rightarrow x \geq \frac{5}{2}$$
Then $$|2x - 5| = 2x - 5$$
So,
$$D = (2x - 5) + 4x^2 - 2x - 13 = 4x^2 - 18$$
Inequality:
$$4x^2 - 18 > 0 \Rightarrow 4x^2 > 18 \Rightarrow x^2 > \frac{9}{2}$$
$$x > \frac{3}{\sqrt{2}} \approx 2.121 \quad \text{or} \quad x < -\frac{3}{\sqrt{2}}$$
Since in this case $$x \geq \frac{5}{2} = 2.5$$, the relevant solution is:
$$x > 2.5$$
- Case 2: $$2x - 5 < 0 \Rightarrow x < \frac{5}{2}$$
Then $$|2x - 5| = -(2x - 5) = 5 - 2x$$
So,
$$D = (5 - 2x) + 4x^2 - 2x - 13 = 4x^2 - 4x - 8$$
Inequality:
$$4x^2 - 4x - 8 > 0$$
Divide both sides by 4:
$$x^2 - x - 2 > 0$$
Factor:
$$(x - 2)(x + 1) > 0$$
The quadratic is positive when:
$$x < -1 \quad \text{or} \quad x > 2$$
But in this case, $$x < 2.5$$, so the solution is:
$$x < -1 \quad \text{or} \quad 2 < x < 2.5$$
8. **Combine the denominator domain conditions:**
$$D > 0$$ when
$$x < -1 \quad \text{or} \quad 2 < x < 2.5 \quad \text{or} \quad x > 2.5$$
Simplify:
$$x < -1 \quad \text{or} \quad x > 2$$
9. **Combine numerator and denominator conditions:**
- Numerator domain: $$x \leq -\frac{7}{3} \approx -2.333 \quad \text{or} \quad x \geq \frac{1}{2} = 0.5$$
- Denominator domain: $$x < -1 \quad \text{or} \quad x > 2$$
10. **Find intersection:**
- For $$x \leq -\frac{7}{3}$$ and $$x < -1$$, the intersection is $$x \leq -\frac{7}{3}$$ (since $$-\frac{7}{3} < -1$$).
- For $$x \geq 0.5$$ and $$x > 2$$, the intersection is $$x > 2$$.
11. **Final domain:**
$$\boxed{(-\infty, -\frac{7}{3}] \cup (2, \infty)}$$
This means the function is defined for all $$x$$ less than or equal to $$-\frac{7}{3}$$ and for all $$x$$ greater than $$2$$.