1.1 Solve For $x$:1.1.1 $(2x + 1)(5 - X) = 0$1.1.2 $x^2 - \frac{2}{3}x - 5 = 0$ (correct To Two Decimal Places)1.1.3 $2x\left(x - \frac{1}{2}\right) \leq 6$1.1.4 \$x + \sqrt{2x - 3} =
Introduction
Equations and inequalities are fundamental concepts in mathematics that play a crucial role in various fields, including physics, engineering, economics, and computer science. In this article, we will delve into the world of equations and inequalities, exploring various techniques for solving them. We will cover the basics of solving linear and quadratic equations, as well as inequalities, and provide step-by-step solutions to several problems.
Solving Linear Equations
Linear equations are equations in which the highest power of the variable is 1. They can be written in the form ax + b = 0, where a and b are constants. To solve a linear equation, we need to isolate the variable x.
1.1 Solve for $x$: (2x + 1)(5 - x) = 0
To solve this equation, we can start by expanding the left-hand side:
(2x + 1)(5 - x) = 2x(5 - x) + 1(5 - x) = 10x - 2x^2 + 5 - x = -2x^2 + 9x + 5
Now, we can set the equation equal to zero:
-2x^2 + 9x + 5 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -2, b = 9, and c = 5. Plugging these values into the formula, we get:
x = (-(9) ± √((9)^2 - 4(-2)(5))) / 2(-2) = (-9 ± √(81 + 40)) / -4 = (-9 ± √121) / -4 = (-9 ± 11) / -4
Simplifying, we get two possible solutions:
x = (-9 + 11) / -4 = 2 / -4 = -1/2 x = (-9 - 11) / -4 = -20 / -4 = 5
Therefore, the solutions to the equation (2x + 1)(5 - x) = 0 are x = -1/2 and x = 5.
1.2 Solve for $x$: x^2 - \frac{2}{3}x - 5 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = -2/3, and c = -5. Plugging these values into the formula, we get:
x = (-(2/3) ± √((-2/3)^2 - 4(1)(-5))) / 2(1) = (-2/3 ± √(4/9 + 20)) / 2 = (-2/3 ± √(4/9 + 180/9)) / 2 = (-2/3 ± √(184/9)) / 2 = (-2/3 ± 2√23/3) / 2
Simplifying, we get two possible solutions:
x = (-2/3 + 2√23/3) / 2 = (√23 - 1)/3 x = (-2/3 - 2√23/3) / 2 = (-√23 - 1)/3
Therefore, the solutions to the equation x^2 - \frac{2}{3}x - 5 = 0 are x = (√23 - 1)/3 and x = (-√23 - 1)/3.
1.3 Solve for $x$: 2x\left(x - \frac{1}{2}\right) \leq 6
To solve this inequality, we can start by expanding the left-hand side:
2x\left(x - \frac{1}{2}\right) = 2x^2 - x
Now, we can set up the inequality:
2x^2 - x ≤ 6
To solve this quadratic inequality, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 2, b = -1, and c = 6. Plugging these values into the formula, we get:
x = (-(1) ± √((-1)^2 - 4(2)(6))) / 2(2) = (-1 ± √(1 - 48)) / 4 = (-1 ± √(-47)) / 4
Since the discriminant is negative, the quadratic equation has no real solutions. Therefore, the inequality 2x\left(x - \frac{1}{2}\right) \leq 6 has no real solutions.
1.4 Solve for $x$: x + \sqrt{2x - 3} = 0
To solve this equation, we can start by isolating the square root:
\sqrt{2x - 3} = -x
Now, we can square both sides:
2x - 3 = x^2
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = 1, and c = -2. Plugging these values into the formula, we get:
x = (-(1) ± √((1)^2 - 4(1)(-2))) / 2(1) = (-1 ± √(1 + 8)) / 2 = (-1 ± √9) / 2 = (-1 ± 3) / 2
Simplifying, we get two possible solutions:
x = (-1 + 3) / 2 = 1 x = (-1 - 3) / 2 = -2
Therefore, the solutions to the equation x + \sqrt{2x - 3} = 0 are x = 1 and x = -2.
Conclusion
In this article, we have explored various techniques for solving linear and quadratic equations, as well as inequalities. We have provided step-by-step solutions to several problems, including solving for x in equations and inequalities. By mastering these techniques, you will be able to solve a wide range of mathematical problems and become proficient in algebra.
References
- [1] "Algebra" by Michael Artin
- [2] "Calculus" by Michael Spivak
- [3] "Linear Algebra" by Jim Hefferon
Further Reading
- [1] "Introduction to Algebra" by Richard Rusczyk
- [2] "Algebra and Trigonometry" by James Stewart
- [3] "Linear Algebra and Its Applications" by Gilbert Strang
Frequently Asked Questions: Solving Equations and Inequalities ====================================================================
Q: What is the difference between a linear equation and a quadratic equation?
A: A linear equation is an equation in which the highest power of the variable is 1, while a quadratic equation is an equation in which the highest power of the variable is 2.
Q: How do I solve a linear equation?
A: To solve a linear equation, you need to isolate the variable x. You can do this by adding or subtracting the same value to both sides of the equation, or by multiplying or dividing both sides of the equation by the same non-zero value.
Q: How do I solve a quadratic equation?
A: To solve a quadratic equation, you can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
where a, b, and c are the coefficients of the quadratic equation.
Q: What is the quadratic formula?
A: The quadratic formula is a formula that allows you to solve a quadratic equation of the form ax^2 + bx + c = 0. It is given by:
x = (-b ± √(b^2 - 4ac)) / 2a
Q: How do I solve an inequality?
A: To solve an inequality, you need to isolate the variable x. You can do this by adding or subtracting the same value to both sides of the inequality, or by multiplying or dividing both sides of the inequality by the same non-zero value.
Q: What is the difference between a strict inequality and a non-strict inequality?
A: A strict inequality is an inequality that is written with a strict symbol, such as < or >, while a non-strict inequality is an inequality that is written with a non-strict symbol, such as ≤ or ≥.
Q: How do I graph a linear equation?
A: To graph a linear equation, you can use the slope-intercept form of the equation, which is given by:
y = mx + b
where m is the slope of the line and b is the y-intercept.
Q: How do I graph a quadratic equation?
A: To graph a quadratic equation, you can use the vertex form of the equation, which is given by:
y = a(x - h)^2 + k
where (h, k) is the vertex of the parabola.
Q: What is the vertex of a parabola?
A: The vertex of a parabola is the point at which the parabola changes direction. It is given by the coordinates (h, k).
Q: How do I find the vertex of a parabola?
A: To find the vertex of a parabola, you can use the formula:
h = -b / 2a k = f(h)
where f(x) is the quadratic function.
Q: What is the domain of a function?
A: The domain of a function is the set of all possible input values for which the function is defined.
Q: What is the range of a function?
A: The range of a function is the set of all possible output values for which the function is defined.
Q: How do I find the domain and range of a function?
A: To find the domain and range of a function, you need to consider the restrictions on the input and output values. You can do this by analyzing the graph of the function or by using algebraic methods.
Conclusion
In this article, we have answered some of the most frequently asked questions about solving equations and inequalities. We have covered topics such as linear and quadratic equations, inequalities, graphing, and domain and range. By mastering these concepts, you will be able to solve a wide range of mathematical problems and become proficient in algebra.