Find The Solution Of The Following Nonlinear Equations.A. ∣ X − 5 ∣ + ∣ X + 3 ∣ = 12 |x-5|+|x+3|=12 ∣ X − 5∣ + ∣ X + 3∣ = 12 B. X + 4 = X − 2 \sqrt{x+4}=x-2 X + 4 ​ = X − 2 C. ∣ 3 X + 4 ∣ − ∣ 5 X − 4 ∣ = 0 |3x+4|-|5x-4|=0 ∣3 X + 4∣ − ∣5 X − 4∣ = 0 D. 3 X + 4 − 4 = 0 \sqrt{3x+4}-4=0 3 X + 4 ​ − 4 = 0 E. X + 3 = 5 \sqrt{x+3}=5 X + 3 ​ = 5 F. ∣ 7 − ( X + 4 ) ∣ + ∣ 3 X − 3 ∣ = 0 |7-(x+4)|+|3x-3|=0 ∣7 − ( X + 4 ) ∣ + ∣3 X − 3∣ = 0

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Introduction

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Nonlinear equations are a fundamental concept in mathematics, and solving them can be a challenging task. In this article, we will explore the solution to several nonlinear equations, including absolute value equations and equations involving square roots. We will use various techniques, such as algebraic manipulation and graphical analysis, to find the solutions to these equations.

Solution to Equation A: $|x-5|+|x+3|=12$

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Equation A involves absolute value, which can be solved by considering different cases based on the sign of the expression inside the absolute value.

Case 1: $x-5 \geq 0$ and $x+3 \geq 0$

In this case, the equation becomes:

$$x-5+x+3=12$$

Simplifying the equation, we get:

$$2x-2=12$$

Adding 2 to both sides, we get:

$$2x=14$$

Dividing both sides by 2, we get:

$$x=7$$

Case 2: $x-5 < 0$ and $x+3 \geq 0$

In this case, the equation becomes:

$$-(x-5)+x+3=12$$

Simplifying the equation, we get:

$$-x+5+x+3=12$$

Combining like terms, we get:

$$8=12$$

This is a contradiction, so there is no solution in this case.

Case 3: $x-5 \geq 0$ and $x+3 < 0$

In this case, the equation becomes:

$$x-5+-(x+3)=12$$

Simplifying the equation, we get:

$$x-5-x-3=12$$

Combining like terms, we get:

$$-8=12$$

This is a contradiction, so there is no solution in this case.

Case 4: $x-5 < 0$ and $x+3 < 0$

In this case, the equation becomes:

$$-(x-5)+-(x+3)=12$$

Simplifying the equation, we get:

$$-x+5-x-3=12$$

Combining like terms, we get:

$$-2x+2=12$$

Subtracting 2 from both sides, we get:

$$-2x=10$$

Dividing both sides by -2, we get:

$$x=-5$$

Therefore, the solutions to Equation A are $x=7$ and $x=-5$.

Solution to Equation B: $\sqrt{x+4}=x-2$

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Equation B involves a square root, which can be solved by squaring both sides of the equation.

Squaring Both Sides

Squaring both sides of the equation, we get:

$$x+4=(x-2)^2$$

Expanding the right-hand side, we get:

$$x+4=x^2-4x+4$$

Rearranging the equation, we get:

$$x^2-5x=0$$

Factoring out $x$, we get:

$$x(x-5)=0$$

This gives us two possible solutions: $x=0$ and $x=5$.

Checking the Solutions

We need to check if these solutions satisfy the original equation.

For $x=0$, we get:

$$\sqrt{0+4}=\sqrt{4}=2$$

$$0-2=-2$$

Since $2 \neq -2$, $x=0$ is not a solution.

For $x=5$, we get:

$$\sqrt{5+4}=\sqrt{9}=3$$

$$5-2=3$$

Since $3=3$, $x=5$ is a solution.

Therefore, the solution to Equation B is $x=5$.

Solution to Equation C: $|3x+4|-|5x-4|=0$

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Equation C involves absolute value, which can be solved by considering different cases based on the sign of the expression inside the absolute value.

Case 1: $3x+4 \geq 0$ and $5x-4 \geq 0$

In this case, the equation becomes:

$$(3x+4)-(5x-4)=0$$

Simplifying the equation, we get:

$$3x+4-5x+4=0$$

Combining like terms, we get:

$$-2x+8=0$$

Subtracting 8 from both sides, we get:

$$-2x=-8$$

Dividing both sides by -2, we get:

$$x=4$$

Case 2: $3x+4 < 0$ and $5x-4 \geq 0$

In this case, the equation becomes:

$$-(3x+4)-(5x-4)=0$$

Simplifying the equation, we get:

$$-3x-4-5x+4=0$$

Combining like terms, we get:

$$-8x=0$$

Dividing both sides by -8, we get:

$$x=0$$

Case 3: $3x+4 \geq 0$ and $5x-4 < 0$

In this case, the equation becomes:

$$(3x+4)+-(5x-4)=0$$

Simplifying the equation, we get:

$$3x+4-5x+4=0$$

Combining like terms, we get:

$$-2x+8=0$$

Subtracting 8 from both sides, we get:

$$-2x=-8$$

Dividing both sides by -2, we get:

$$x=4$$

Case 4: $3x+4 < 0$ and $5x-4 < 0$

In this case, the equation becomes:

$$-(3x+4)+-(5x-4)=0$$

Simplifying the equation, we get:

$$-3x-4-5x+4=0$$

Combining like terms, we get:

$$-8x=0$$

Dividing both sides by -8, we get:

$$x=0$$

Therefore, the solutions to Equation C are $x=0$ and $x=4$.

Solution to Equation D: $\sqrt{3x+4}-4=0$

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Equation D involves a square root, which can be solved by isolating the square root and then squaring both sides of the equation.

Isolating the Square Root

Adding 4 to both sides of the equation, we get:

$$\sqrt{3x+4}=4$$

Squaring Both Sides

Squaring both sides of the equation, we get:

$$3x+4=16$$

Subtracting 4 from both sides, we get:

$$3x=12$$

Dividing both sides by 3, we get:

$$x=4$$

Therefore, the solution to Equation D is $x=4$.

Solution to Equation E: $\sqrt{x+3}=5$

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Equation E involves a square root, which can be solved by isolating the square root and then squaring both sides of the equation.

Isolating the Square Root

Squaring both sides of the equation, we get:

$$x+3=25$$

Subtracting 3 from both sides, we get:

$$x=22$$

Therefore, the solution to Equation E is $x=22$.

Solution to Equation F: $|7-(x+4)|+|3x-3|=0$

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Equation F involves absolute value, which can be solved by considering different cases based on the sign of the expression inside the absolute value.

Case 1: $7-(x+4) \geq 0$ and $3x-3 \geq 0$

In this case, the equation becomes:

$$(7-(x+4))+(3x-3)=0$$

Simplifying the equation, we get:

$$7-x-4+3x-3=0$$

Combining like terms, we get:

$$2x+0=0$$

Subtracting 0 from both sides, we get:

$$2x=0$$

Dividing both sides by 2, we get:

$$x=0$$

Case 2: $7-(x+4) < 0$ and $3x-3 \geq 0$

In this case, the equation becomes:

$$-(7-(x+4))+(3x-3)=0$$

Simplifying the equation, we get:

$$-(7-x-4)+(3x-3)=0$$

Combining like terms, we get:

$$-3+x+3x-3=0$$

Combining like terms, we get:

$$4x=0$$

Dividing both sides by 4, we get:

$$x=0$$

Case 3: $7-(x+4) \geq 0$ and $3x-3 < 0$

In this case, the equation becomes:

(7-(x+4))+<br/> # **Frequently Asked Questions: Solving Nonlinear Equations** ===========================================================

Q: What is a nonlinear equation?

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A: A nonlinear equation is an equation that cannot be written in the form of a linear equation, which is an equation of the form $ax + b = 0$, where $a$ and $b$ are constants. Nonlinear equations can involve absolute value, square roots, and other nonlinear functions.

Q: How do I solve a nonlinear equation?

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A: Solving a nonlinear equation can be challenging, but there are several techniques that can be used. These include:

• Graphical analysis: Plotting the equation on a graph and finding the points of intersection with the x-axis.
• Algebraic manipulation: Rearranging the equation to isolate the nonlinear term and then solving for the variable.
• Numerical methods: Using numerical methods, such as the Newton-Raphson method, to approximate the solution.

Q: What is the difference between a linear and nonlinear equation?

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A: The main difference between a linear and nonlinear equation is the form of the equation. A linear equation can be written in the form of $ax + b = 0$, where $a$ and $b$ are constants. A nonlinear equation, on the other hand, cannot be written in this form and may involve absolute value, square roots, and other nonlinear functions.

Q: Can all nonlinear equations be solved?

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A: Unfortunately, not all nonlinear equations can be solved. Some nonlinear equations may have no solution, while others may have multiple solutions. In some cases, the equation may be too complex to solve analytically and numerical methods may be required.

Q: How do I know if a nonlinear equation has a solution?

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A: To determine if a nonlinear equation has a solution, you can try the following:

• Check for contradictions: If the equation involves absolute value or square roots, check for contradictions by squaring both sides or isolating the nonlinear term.
• Graphical analysis: Plot the equation on a graph and check for points of intersection with the x-axis.
• Numerical methods: Use numerical methods, such as the Newton-Raphson method, to approximate the solution.

Q: What are some common types of nonlinear equations?

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A: Some common types of nonlinear equations include:

• Absolute value equations: Equations that involve absolute value, such as $|x| = 5$.
• Square root equations: Equations that involve square roots, such as $\sqrt{x} = 5$.
• Polynomial equations: Equations that involve polynomials, such as $x^2 + 2x + 1 = 0$.
• Trigonometric equations: Equations that involve trigonometric functions, such as $\sin(x) = 0$.

Q: How do I solve a nonlinear equation with absolute value?

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A: To solve a nonlinear equation with absolute value, you can try the following:

• Isolate the absolute value term: Rearrange the equation to isolate the absolute value term.
• Consider different cases: Consider different cases based on the sign of the expression inside the absolute value.
• Solve for the variable: Solve for the variable in each case.

Q: How do I solve a nonlinear equation with a square root?

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A: To solve a nonlinear equation with a square root, you can try the following:

• Isolate the square root term: Rearrange the equation to isolate the square root term.
• Square both sides: Square both sides of the equation to eliminate the square root.
• Solve for the variable: Solve for the variable.

Q: What are some common mistakes to avoid when solving nonlinear equations?

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A: Some common mistakes to avoid when solving nonlinear equations include:

• Not checking for contradictions: Failing to check for contradictions when solving absolute value or square root equations.
• Not considering different cases: Failing to consider different cases when solving absolute value or square root equations.
• Not using numerical methods: Failing to use numerical methods when solving complex nonlinear equations.

By following these tips and avoiding common mistakes, you can increase your chances of successfully solving nonlinear equations.