Select Whether Each Equation Has No Real Solutions, One Real Solution, Or Infinitely Many Real Solutions.1. X + 7 = 2 \sqrt{x} + 7 = 2 X ​ + 7 = 2 - [ ] No Real Solutions - [ ] One Real Solution - [ ] Infinitely Many Real Solutions2. $\frac{15}{x+1} =

Introduction

Equations are a fundamental concept in mathematics, and understanding the nature of their solutions is crucial for solving various mathematical problems. In this article, we will explore the concept of real solutions in equations and learn how to determine whether an equation has no real solutions, one real solution, or infinitely many real solutions.

What are Real Solutions?

A real solution is a value of the variable that makes the equation true. In other words, it is a value that satisfies the equation. Real solutions can be either rational or irrational numbers.

Understanding the Nature of Solutions

To determine the nature of the solutions of an equation, we need to analyze its structure. We can use various techniques such as factoring, quadratic formula, and graphing to understand the behavior of the equation.

Equation 1: $\sqrt{x} + 7 = 2$

Let's start with the first equation: $\sqrt{x} + 7 = 2$. To solve this equation, we need to isolate the square root term.

Step 1: Subtract 7 from both sides

$\sqrt{x} + 7 - 7 = 2 - 7$

$\sqrt{x} = -5$

Step 2: Square both sides

$(\sqrt{x})^2 = (-5)^2$

$x = 25$

However, we need to check if this solution is valid. Since the square root of a negative number is not a real number, we need to consider the domain of the square root function.

Domain of the Square Root Function

The domain of the square root function is all non-negative real numbers. In other words, $x \geq 0$. Since $x = 25$ is a positive number, it is a valid solution.

Conclusion

The equation $\sqrt{x} + 7 = 2$ has one real solution, which is $x = 25$.

Equation 2: $\frac{15}{x+1} = 2$

Let's move on to the second equation: $\frac{15}{x+1} = 2$. To solve this equation, we need to isolate the variable.

Step 1: Multiply both sides by $x+1$

$\frac{15}{x+1} \cdot (x+1) = 2 \cdot (x+1)$

$15 = 2x + 2$

Step 2: Subtract 2 from both sides

$15 - 2 = 2x + 2 - 2$

$13 = 2x$

Step 3: Divide both sides by 2

$\frac{13}{2} = \frac{2x}{2}$

$\frac{13}{2} = x$

Conclusion

The equation $\frac{15}{x+1} = 2$ has one real solution, which is $x = \frac{13}{2}$.

Equation 3: $x^2 + 4 = 0$

Let's consider the third equation: $x^2 + 4 = 0$. To solve this equation, we need to isolate the variable.

Step 1: Subtract 4 from both sides

$x^2 + 4 - 4 = 0 - 4$

$x^2 = -4$

Step 2: Take the square root of both sides

$\sqrt{x^2} = \sqrt{-4}$

$x = \pm \sqrt{-4}$

Conclusion

The equation $x^2 + 4 = 0$ has no real solutions, since the square root of a negative number is not a real number.

Equation 4: $x^2 - 4 = 0$

Let's consider the fourth equation: $x^2 - 4 = 0$. To solve this equation, we need to isolate the variable.

Step 1: Add 4 to both sides

$x^2 - 4 + 4 = 0 + 4$

$x^2 = 4$

Step 2: Take the square root of both sides

$\sqrt{x^2} = \sqrt{4}$

$x = \pm 2$

Conclusion

The equation $x^2 - 4 = 0$ has two real solutions, which are $x = 2$ and $x = -2$.

Equation 5: $\sqrt{x} = 0$

Let's consider the fifth equation: $\sqrt{x} = 0$. To solve this equation, we need to isolate the variable.

Step 1: Square both sides

$(\sqrt{x})^2 = 0^2$

$x = 0$

Conclusion

The equation $\sqrt{x} = 0$ has one real solution, which is $x = 0$.

Equation 6: $x^2 + 2x + 1 = 0$

Let's consider the sixth equation: $x^2 + 2x + 1 = 0$. To solve this equation, we need to factor it.

Step 1: Factor the equation

$(x + 1)^2 = 0$

Step 2: Take the square root of both sides

$\sqrt{(x + 1)^2} = \sqrt{0}$

$x + 1 = 0$

Step 3: Subtract 1 from both sides

$x + 1 - 1 = 0 - 1$

$x = -1$

Conclusion

The equation $x^2 + 2x + 1 = 0$ has one real solution, which is $x = -1$.

Equation 7: $x^2 - 4x + 4 = 0$

Let's consider the seventh equation: $x^2 - 4x + 4 = 0$. To solve this equation, we need to factor it.

Step 1: Factor the equation

$(x - 2)^2 = 0$

Step 2: Take the square root of both sides

$\sqrt{(x - 2)^2} = \sqrt{0}$

$x - 2 = 0$

Step 3: Add 2 to both sides

$x - 2 + 2 = 0 + 2$

$x = 2$

Conclusion

The equation $x^2 - 4x + 4 = 0$ has one real solution, which is $x = 2$.

Equation 8: $x^2 + 1 = 0$

Let's consider the eighth equation: $x^2 + 1 = 0$. To solve this equation, we need to isolate the variable.

Step 1: Subtract 1 from both sides

$x^2 + 1 - 1 = 0 - 1$

$x^2 = -1$

Step 2: Take the square root of both sides

$\sqrt{x^2} = \sqrt{-1}$

$x = \pm i$

Conclusion

The equation $x^2 + 1 = 0$ has no real solutions, since the square root of a negative number is not a real number.

Equation 9: $\frac{1}{x} = 2$

Let's consider the ninth equation: $\frac{1}{x} = 2$. To solve this equation, we need to isolate the variable.

Step 1: Multiply both sides by $x$

$\frac{1}{x} \cdot x = 2 \cdot x$

$1 = 2x$

Step 2: Divide both sides by 2

$\frac{1}{2} = \frac{2x}{2}$

$\frac{1}{2} = x$

Conclusion

The equation $\frac{1}{x} = 2$ has one real solution, which is $x = \frac{1}{2}$.

Equation 10: $x^2 - 3x + 2 = 0$

Let's consider the tenth equation: $x^2 - 3x + 2 = 0$. To solve this equation, we need to factor it.

Step 1: Factor the equation

$(x - 2)(x - 1) = 0$

Step 2: Set each factor equal to 0

$x - 2 = 0$ or $x - 1 = 0$

Step 3: Solve for $x$

$x = 2$ or $x = 1$

Conclusion

The equation $x^2 - 3x + 2 = 0$ has two real solutions, which are $x = 2$ and $x = 1$.

Conclusion

Q&A: Real Solutions in Equations

Q: What is a real solution in an equation?

A: A real solution is a value of the variable that makes the equation true. In other words, it is a value that satisfies the equation.

Q: How do I determine whether an equation has no real solutions, one real solution, or infinitely many real solutions?

A: To determine the nature of the solutions of an equation, you need to analyze its structure. You can use various techniques such as factoring, quadratic formula, and graphing to understand the behavior of the equation.

Q: What is the difference between a rational and an irrational solution?

A: A rational solution is a solution that can be expressed as a ratio of two integers, while an irrational solution is a solution that cannot be expressed as a ratio of two integers.

Q: How do I find the real solutions of a quadratic equation?

A: To find the real solutions of a quadratic equation, you can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. You can also factor the equation, if possible.

Q: What is the significance of the discriminant in a quadratic equation?

A: The discriminant is the expression under the square root in the quadratic formula: $b^2 - 4ac$. If the discriminant is positive, the equation has two real solutions. If the discriminant is zero, the equation has one real solution. If the discriminant is negative, the equation has no real solutions.

Q: How do I determine whether an equation has infinitely many real solutions?

A: An equation has infinitely many real solutions if it is an identity, meaning that it is true for all values of the variable.

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

A: A linear equation is an equation of the form $ax + b = 0$, while a quadratic equation is an equation of the form $ax^2 + bx + c = 0$.

Q: How do I solve a linear equation?

A: To solve a linear equation, you can use the following steps:

1. Add or subtract the same value to both sides of the equation to isolate the variable.
2. Multiply or divide both sides of the equation by the same value to isolate the variable.

Q: How do I solve a quadratic equation?

A: To solve a quadratic equation, you can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. You can also factor the equation, if possible.

Q: What is the difference between a rational and an irrational solution?

A: A rational solution is a solution that can be expressed as a ratio of two integers, while an irrational solution is a solution that cannot be expressed as a ratio of two integers.

Q: How do I determine whether an equation has a rational or irrational solution?

A: To determine whether an equation has a rational or irrational solution, you need to analyze the equation and determine whether the solution can be expressed as a ratio of two integers.

Q: What is the significance of the domain of an equation?

A: The domain of an equation is the set of all possible values of the variable. It is essential to consider the domain of an equation when solving it, as it can affect the validity of the solution.

Q: How do I determine the domain of an equation?

A: To determine the domain of an equation, you need to analyze the equation and determine the set of all possible values of the variable.

Conclusion

In this article, we have explored the concept of real solutions in equations and answered some frequently asked questions about real solutions. We have seen that real solutions can be rational or irrational, and that the nature of the solutions of an equation depends on its structure. We have also seen that some equations have no real solutions, while others have one or two real solutions.