Question

Solution set of the inequality $\sqrt{x^2-x-6} < 2x-3$ is

• A. $x \in [-1, \frac{\sqrt{5}-1}{2})$
• B. $x \in [\frac{16+\sqrt{7}}{2}, 10]$
• C. $x \in [3, \infty)$

Answer 1

Answer: (C)

$x \in [3, \infty)$

Answer 2

To solve the inequality $\sqrt{x^2-x-6} < 2x-3$, we need to consider three main conditions:

1. Domain of the square root:

For $\sqrt{x^2-x-6}$ to be defined in real numbers, the expression under the square root must be non-negative:

$x^2-x-6 \ge 0$

Factor the quadratic: $(x-3)(x+2) \ge 0$

The roots are $x=3$ and $x=-2$.

This inequality holds when $x \le -2$ or $x \ge 3$.

So, $x \in (-\infty, -2] \cup [3, \infty)$. (Condition 1)

2. Right Hand Side (RHS) must be positive:

For the inequality $\sqrt{A} < B$ to have a solution, $B$ must be positive, since the square root of a real number is always non-negative.

$2x-3 > 0$

$2x > 3$

$x > \frac{3}{2}$ (Condition 2)

3. Squaring both sides:

Since both sides of the inequality are non-negative (from Condition 2, $2x-3 > 0$, and $\sqrt{x^2-x-6} \ge 0$), we can square both sides without changing the direction of the inequality:

$(\sqrt{x^2-x-6})^2 < (2x-3)^2$

$x^2-x-6 < 4x^2 - 12x + 9$

Rearrange the terms to form a quadratic inequality:

$0 < 4x^2 - x^2 - 12x + x + 9 + 6$

$0 < 3x^2 - 11x + 15$

$3x^2 - 11x + 15 > 0$ (Condition 3)

To analyze Condition 3, let's find the discriminant ($\Delta$) of the quadratic $3x^2 - 11x + 15$:

$\Delta = b^2 - 4ac = (-11)^2 - 4(3)(15)$

$\Delta = 121 - 180$

$\Delta = -59$

Since the discriminant is negative ($\Delta < 0$) and the leading coefficient is positive ($a=3 > 0$), the quadratic $3x^2 - 11x + 15$ is always positive for all real values of $x$.

So, Condition 3 is true for all $x \in \mathbb{R}$.

4. Combine all conditions:

We need to find the intersection of Condition 1, Condition 2, and Condition 3.

Condition 1: $x \in (-\infty, -2] \cup [3, \infty)$

Condition 2: $x \in (\frac{3}{2}, \infty)$

Condition 3: $x \in \mathbb{R}$

Let's find the intersection of Condition 1 and Condition 2:

The interval $(-\infty, -2]$ does not overlap with $(\frac{3}{2}, \infty)$ because $-2 < \frac{3}{2}$.

The interval $[3, \infty)$ overlaps with $(\frac{3}{2}, \infty)$. Since $3 > \frac{3}{2}$, the intersection of $[3, \infty)$ and $(\frac{3}{2}, \infty)$ is $[3, \infty)$.

Since Condition 3 is true for all real $x$, it does not further restrict the solution set.

Therefore, the solution set for the inequality is $x \in [3, \infty)$.