Solving Inequalities With Square Roots In R

Hey guys! Today, we're diving into how to solve inequalities involving square roots using R. These types of problems might seem tricky at first, but with a systematic approach and a bit of R magic, you'll be able to tackle them like a pro. We'll be covering three specific inequalities, breaking down each step to make sure you understand the logic behind it.

a) Solving $\sqrt{1-x^2} < 2x$ in R

Let's kick things off with our first inequality: $\sqrt{1-x^2} < 2x$. When dealing with square roots, the initial crucial step is to ensure that the expression inside the square root is non-negative. Additionally, we need to consider the sign of the other side of the inequality to avoid any pitfalls. Here’s a detailed breakdown:

1. Determine the Domain of the Square Root

The expression inside the square root must be greater than or equal to zero. Thus, we have:

$1 - x^2 \ge 0$

This implies:

$x^2 \le 1$

Which gives us the domain:

$-1 \le x \le 1$

2. Consider the Sign of $2x$

Since the square root is always non-negative, $2x$ must also be positive for the inequality to hold. Therefore:

$2x > 0$

Which means:

$x > 0$

Combining this with our domain, we now have:

$0 < x \le 1$

3. Square Both Sides

Now that we've established the necessary conditions, we can safely square both sides of the inequality:

$(\sqrt{1-x^2})^2 < (2x)^2$

This simplifies to:

$1 - x^2 < 4x^2$

4. Rearrange and Solve the Inequality

Rearrange the inequality to get:

$5x^2 > 1$

Which implies:

$x^2 > \frac{1}{5}$

Taking the square root of both sides:

$x > \frac{1}{\sqrt{5}}$ or $x < -\frac{1}{\sqrt{5}}$

5. Combine All Conditions

We need to combine the following conditions:

• $0 < x \le 1$
• $x > \frac{1}{\sqrt{5}}$ or $x < -\frac{1}{\sqrt{5}}$

Since $x$ must be positive, we only consider $x > \frac{1}{\sqrt{5}}$. Therefore, the solution is:

$\frac{1}{\sqrt{5}} < x \le 1$

In R, you can verify this solution or find it numerically. For example, you can define a function and check its values over the interval. This ensures that we've correctly solved the inequality. The solution set represents all the x values that satisfy the given inequality, making it a fundamental aspect of mathematical analysis.

b) Solving $2x + 1

greatercurlyeq \sqrt{x^2 +3}$ in R

Next up, let's tackle the inequality $2x + 1 \le \sqrt{x^2 + 3}$. This one is a bit different, but we'll approach it systematically, making sure we account for all the necessary conditions. First, keep in mind that whatever is under the square root must be non-negative, which will define the domain, and that the result of the square root is non-negative.

1. Determine the Domain

Since $x^2 + 3$ is always positive for any real number $x$, the domain is all real numbers. That's one less thing to worry about!

2. Consider the Sign of $2x + 1$

In this case, $2x + 1$ can be negative, zero, or positive. If $2x + 1$ is negative, the inequality holds true since the square root is always non-negative. So, we have:

If $2x + 1 < 0$, then $x < -\frac{1}{2}$.

3. Square Both Sides (When $2x + 1 \ge 0$)

When $2x + 1 \ge 0$, i.e., $x \ge -\frac{1}{2}$, we can square both sides of the inequality:

$(2x + 1)^2 \le x^2 + 3$

Expanding and simplifying:

$4x^2 + 4x + 1 \le x^2 + 3$

$3x^2 + 4x - 2 \le 0$

4. Solve the Quadratic Inequality

To solve this quadratic inequality, we first find the roots of the quadratic equation $3x^2 + 4x - 2 = 0$. Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-4 \pm \sqrt{4^2 - 4(3)(-2)}}{2(3)}$

$x = \frac{-4 \pm \sqrt{16 + 24}}{6}$

$x = \frac{-4 \pm \sqrt{40}}{6}$

$x = \frac{-4 \pm 2\sqrt{10}}{6}$

$x = \frac{-2 \pm \sqrt{10}}{3}$

So the roots are $x_1 = \frac{-2 - \sqrt{10}}{3}$ and $x_2 = \frac{-2 + \sqrt{10}}{3}$.

Since the coefficient of $x^2$ is positive, the parabola opens upwards. Therefore, the inequality $3x^2 + 4x - 2 \le 0$ holds between the roots:

$\frac{-2 - \sqrt{10}}{3} \le x \le \frac{-2 + \sqrt{10}}{3}$

5. Combine All Conditions

We need to combine the following conditions:

• $x < -\frac{1}{2}$
• $\frac{-2 - \sqrt{10}}{3} \le x \le \frac{-2 + \sqrt{10}}{3}$ and $x \ge -\frac{1}{2}$

For the second condition, we need to consider only the part where $x \ge -\frac{1}{2}$. Thus,

$-\frac{1}{2} \le x \le \frac{-2 + \sqrt{10}}{3}$

Combining both conditions, the solution is:

$x < -\frac{1}{2}$ or $-\frac{1}{2} \le x \le \frac{-2 + \sqrt{10}}{3}$

This simplifies to:

$x \le \frac{-2 + \sqrt{10}}{3}$

In R, you can again verify this solution by plotting the inequality and checking values within this range. Remember, accurate solutions are achieved when all initial considerations are accounted for, such as the domain and the sign of the expressions involved. This thorough approach ensures that the solution set is complete and correct, a vital practice in solving mathematical problems.

c) Solving $\sqrt{x^2 - 8x + 15} \ge 3x - 2$ in R

Finally, let's tackle $\sqrt{x^2 - 8x + 15} \ge 3x - 2$. This inequality brings another layer of complexity, but don't worry, we'll break it down step by step.

1. Determine the Domain

The expression inside the square root must be non-negative:

$x^2 - 8x + 15 \ge 0$

Factor the quadratic:

$(x - 3)(x - 5) \ge 0$

This inequality holds when $x \le 3$ or $x \ge 5$.

2. Consider the Sign of $3x - 2$

If $3x - 2 < 0$, i.e., $x < \frac{2}{3}$, the inequality holds true because the square root is always non-negative. So we have to consider it only when $x \le 3$.

3. Square Both Sides (When $3x - 2 \ge 0$)

When $3x - 2 \ge 0$, i.e., $x \ge \frac{2}{3}$, we can square both sides of the inequality:

$x^2 - 8x + 15 \ge (3x - 2)^2$

Expanding and simplifying:

$x^2 - 8x + 15 \ge 9x^2 - 12x + 4$

$0 \ge 8x^2 - 4x - 11$

4. Solve the Quadratic Inequality

To solve $8x^2 - 4x - 11 \le 0$, we first find the roots of the quadratic equation $8x^2 - 4x - 11 = 0$. Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{4 \pm \sqrt{(-4)^2 - 4(8)(-11)}}{2(8)}$

$x = \frac{4 \pm \sqrt{16 + 352}}{16}$

$x = \frac{4 \pm \sqrt{368}}{16}$

$x = \frac{4 \pm 4\sqrt{23}}{16}$

$x = \frac{1 \pm \sqrt{23}}{4}$

So the roots are $x_1 = \frac{1 - \sqrt{23}}{4}$ and $x_2 = \frac{1 + \sqrt{23}}{4}$.

Since the coefficient of $x^2$ is positive, the parabola opens upwards. Therefore, the inequality $8x^2 - 4x - 11 \le 0$ holds between the roots:

$\frac{1 - \sqrt{23}}{4} \le x \le \frac{1 + \sqrt{23}}{4}$

5. Combine All Conditions

We need to combine the following conditions:

• $x \le 3$ or $x \ge 5$
• $x < \frac{2}{3}$
• $x \ge \frac{2}{3}$ and $\frac{1 - \sqrt{23}}{4} \le x \le \frac{1 + \sqrt{23}}{4}$

From the first condition and $x < \frac{2}{3}$, we have $x \le 3$ and $x < \frac{2}{3}$, so $x < \frac{2}{3}$.

From the first and third conditions, we have $x \ge \frac{2}{3}$, $x \le 3$ or $x \ge 5$, and $\frac{1 - \sqrt{23}}{4} \le x \le \frac{1 + \sqrt{23}}{4}$.

Since $\frac{1 - \sqrt{23}}{4} \approx -0.9$ and $\frac{1 + \sqrt{23}}{4} \approx 1.4$, we only consider the interval $\frac{2}{3} \le x \le \frac{1 + \sqrt{23}}{4}$. Combining these, the solution is:

$x < \frac{2}{3}$ or $\frac{2}{3} \le x \le \frac{1 + \sqrt{23}}{4}$

This simplifies to:

$x \le \frac{1 + \sqrt{23}}{4}$

And considering the domain:

$x \le 3$ or $x \ge 5$, the solution is $x \in (-\infty, \frac{1 + \sqrt{23}}{4}] \cup [5, + \infty)$.

This thorough approach guarantees an accurate solution, combining algebraic methods with an understanding of inequalities. Remember to verify your solutions using R or other tools to ensure correctness.

Solving inequalities with square roots can be a bit of a puzzle, but with a clear, step-by-step approach, you can definitely master them. Keep practicing, and you'll become more confident in your abilities. Good luck, and have fun solving!