The Equations $y = 2x - 5$ And $y = \sqrt{3x} - 1$ Are Graphed On The Coordinate Grid.How Many Real Solutions Does The Equation $\sqrt{3z} - 1 = 22 - 5$ Have?

Introduction

In mathematics, equations are used to represent relationships between variables. When solving equations, we often encounter real solutions, which are values that satisfy the equation. In this article, we will analyze the equation $\sqrt{3z} - 1 = 22 - 5$ and determine the number of real solutions it has.

Understanding the Equation

The given equation is $\sqrt{3z} - 1 = 22 - 5$. To simplify this equation, we can start by evaluating the right-hand side:

$$\sqrt{3z} - 1 = 17$$

Next, we can add 1 to both sides of the equation to isolate the square root term:

$$\sqrt{3z} = 18$$

Solving for z

To solve for z, we can square both sides of the equation:

$$3z = 18^2$$

$$3z = 324$$

Now, we can divide both sides of the equation by 3 to solve for z:

$$z = \frac{324}{3}$$

$$z = 108$$

Analyzing the Solution

At first glance, it may seem like the equation has only one real solution, which is z = 108. However, we need to consider the domain of the square root function. The square root function is defined only for non-negative values, so we need to check if the value of z is non-negative.

Checking the Domain

Since z = 108 is a positive value, it satisfies the domain of the square root function. Therefore, the equation $\sqrt{3z} - 1 = 22 - 5$ has one real solution, which is z = 108.

Conclusion

In conclusion, the equation $\sqrt{3z} - 1 = 22 - 5$ has one real solution, which is z = 108. This solution satisfies the domain of the square root function, and it is a valid solution to the equation.

The Graphical Representation

To visualize the solution, we can graph the equation on a coordinate grid. The graph of the equation $\sqrt{3z} - 1 = 22 - 5$ is a horizontal line at y = 18. The graph of the equation $y = 2x - 5$ is a line with a slope of 2 and a y-intercept of -5. The graph of the equation $y = \sqrt{3x} - 1$ is a curve that opens upwards.

The Intersection Points

To find the intersection points of the two graphs, we can set the two equations equal to each other:

$$2x - 5 = \sqrt{3x} - 1$$

Solving for x, we get:

$$2x - 5 + 1 = \sqrt{3x}$$

$$2x - 4 = \sqrt{3x}$$

Squaring both sides of the equation, we get:

$$(2x - 4)^2 = 3x$$

Expanding the left-hand side of the equation, we get:

$$4x^2 - 16x + 16 = 3x$$

Rearranging the terms, we get:

$$4x^2 - 19x + 16 = 0$$

Solving for x, we get:

$$x = \frac{19 \pm \sqrt{(-19)^2 - 4(4)(16)}}{2(4)}$$

$$x = \frac{19 \pm \sqrt{361 - 256}}{8}$$

$$x = \frac{19 \pm \sqrt{105}}{8}$$

Since the value of x is not an integer, the graphs of the two equations do not intersect at any integer value of x.

The Number of Real Solutions

Since the graphs of the two equations do not intersect at any integer value of x, the equation $\sqrt{3z} - 1 = 22 - 5$ has only one real solution, which is z = 108.

The Final Answer

In conclusion, the equation $\sqrt{3z} - 1 = 22 - 5$ has one real solution, which is z = 108. This solution satisfies the domain of the square root function, and it is a valid solution to the equation.

The Importance of Real Solutions

Real Solutions in Mathematics

Real solutions are an essential concept in mathematics, as they represent the values that satisfy an equation. In this article, we analyzed the equation $\sqrt{3z} - 1 = 22 - 5$ and determined the number of real solutions it has.

Real Solutions in Real-World Applications

Real solutions have numerous applications in real-world problems. For example, in physics, real solutions are used to model the motion of objects. In engineering, real solutions are used to design and optimize systems.

The Significance of Real Solutions

Real solutions are significant because they provide a way to model and analyze complex systems. They are used in a wide range of fields, including physics, engineering, economics, and computer science.

The Future of Real Solutions

Advances in Mathematics

Advances in mathematics have led to a better understanding of real solutions. New techniques and tools have been developed to analyze and solve equations with real solutions.

Applications of Real Solutions

The applications of real solutions continue to grow. New fields and industries are emerging that rely on real solutions to model and analyze complex systems.

The Importance of Real Solutions in the Future

Real solutions will continue to play a crucial role in mathematics and its applications. As new technologies and techniques emerge, the importance of real solutions will only continue to grow.

Conclusion

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Q: What is a real solution?

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

Q: How do you find real solutions?

A: To find real solutions, you need to solve the equation. This involves using algebraic techniques, such as adding, subtracting, multiplying, and dividing, to isolate the variable.

Q: What are some common types of equations that have real solutions?

A: Some common types of equations that have real solutions include linear equations, quadratic equations, and polynomial equations.

Q: Can you give an example of a linear equation with a real solution?

A: Yes, here is an example of a linear equation with a real solution:

2x + 3 = 5

To solve for x, you can subtract 3 from both sides of the equation:

2x = 5 - 3

2x = 2

Next, you can divide both sides of the equation by 2:

x = 2/2

x = 1

So, the real solution to the equation 2x + 3 = 5 is x = 1.

Q: Can you give an example of a quadratic equation with a real solution?

A: Yes, here is an example of a quadratic equation with a real solution:

x^2 + 4x + 4 = 0

To solve for x, you can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 1, b = 4, and c = 4. Plugging these values into the formula, you get:

x = (-(4) ± √((4)^2 - 4(1)(4))) / 2(1)

x = (-4 ± √(16 - 16)) / 2

x = (-4 ± √0) / 2

x = (-4 ± 0) / 2

x = -4/2

x = -2

So, the real solution to the equation x^2 + 4x + 4 = 0 is x = -2.

Q: Can you give an example of a polynomial equation with a real solution?

A: Yes, here is an example of a polynomial equation with a real solution:

x^3 + 2x^2 + x + 1 = 0

To solve for x, you can use the rational root theorem, which states that if a rational number p/q is a root of the polynomial, then p must be a factor of the constant term and q must be a factor of the leading coefficient.

In this case, the constant term is 1, and the leading coefficient is 1. The only factors of 1 are ±1, so the only possible rational roots are ±1.

To test these possible roots, you can plug them into the polynomial:

(-1)^3 + 2(-1)^2 + (-1) + 1 = -1 + 2 - 1 + 1 = 1

(1)^3 + 2(1)^2 + (1) + 1 = 1 + 2 + 1 + 1 = 5

Since neither of these values is equal to 0, the polynomial has no rational roots. However, we can still try to find a real solution by using numerical methods, such as the Newton-Raphson method.

Q: What are some common mistakes to avoid when finding real solutions?

A: Some common mistakes to avoid when finding real solutions include:

• Not checking the domain of the equation
• Not using the correct algebraic techniques
• Not checking for extraneous solutions
• Not using numerical methods when necessary

Q: How can you check the domain of an equation?

A: To check the domain of an equation, you need to determine the values of the variable that make the equation true. For example, if the equation contains a square root, you need to check that the value inside the square root is non-negative.

Q: How can you check for extraneous solutions?

A: To check for extraneous solutions, you need to plug the solution back into the original equation and check that it is true. If the solution is not true, then it is an extraneous solution.

Q: What are some common numerical methods for finding real solutions?

A: Some common numerical methods for finding real solutions include:

• The Newton-Raphson method
• The bisection method
• The secant method

These methods involve using iterative techniques to approximate the real solution.

Q: How can you use the Newton-Raphson method to find a real solution?

A: To use the Newton-Raphson method, you need to start with an initial guess for the real solution. Then, you can use the formula:

x_n+1 = x_n - f(x_n) / f'(x_n)

where x_n is the current estimate of the real solution, f(x_n) is the value of the function at x_n, and f'(x_n) is the derivative of the function at x_n.

You can repeat this process until the estimate converges to the real solution.

Q: What are some common applications of real solutions?

A: Some common applications of real solutions include:

• Modeling the motion of objects in physics
• Designing and optimizing systems in engineering
• Analyzing data in economics and finance
• Solving problems in computer science and mathematics

Real solutions are an essential tool for solving problems in a wide range of fields.