Question

1. # Using The Quadratic Formula To Solve 4X2 – 3X + 9 = 2X + 1, What Are The Values Of X?

The Quadratic Formula is a powerful tool that can help algebra students solve quadratic equations. In this article, we will discuss how to use the Quadratic Formula to solve 4×2 – 3x + 9 = 2x + 1. We will also explore how to calculate the values of X, so that you can get an accurate solution for your equation.

## What is the Quadratic Formula?

The Quadratic Formula is a mathematical formula used to solve for the roots of a quadratic equation. The quadratic equation is any equation that can be written in the form: ax^2 + bx + c = 0. where a, b, and c are real numbers and x is an unknown. The Quadratic Formula can be used to solve for any one of the roots of the equation. In this case, the two roots are: x = (-b +/- sqrt(b^2-4ac)) / (2a).

## How to Use the Quadratic Formula

To use the Quadratic Formula, we need to first identify what a, b, and c are in the equation. In this equation, a = 1, b = -1, and c = 1. Now that we know what a, b, and c are, we can plug them into the Quadratic Formula:

-b ± √(b^2-4ac)
2a

plugged in for our equation becomes:

1 ± √(1-(4(1)(1))
2(1)

Which simplifies to:

1 ± √(-3)
2

## The Quadratic Formula and 4X2 – 3X + 9 = 2X + 1

The quadratic equation is a very useful mathematical tool that can be used to solve a variety of problems. The quadratic equation can be used to solve for the roots of a polynomial, the zeroes of a function, or the points of intersection of two lines. The quadratic equation can also be used to find the maximum or minimum value of a function. In this blog article, we will be using the quadratic equation to solve for the values of x in the equation 4×2 – 3x + 9 = 2x + 1.

To solve for the roots of a quadratic equation, we need to use the quadratic formula. The quadratic formula is as follows: x = -b ± √(b2 – 4ac) / 2a. In our equation, 4×2 – 3x + 9 = 2x + 1, we have a = 4, b = -3, and c = 9. Plugging these values into our quadratic formula, we get: x = -(-3) ± √((-3)2 – 4(4)(9)) / 2(4). Simplifying this expression, we get: x = 3 ± √(9 – 144) / 8. Since we are looking for the values of x that make our equation true, we need to find the two values of x that make our simplified expression equal to 0. To do this, we can set our equation equal to 0 and solve for x. We get: 3 ± √(9 – 144) / 8 = 0. Simplifying this equation, we get: 3 ± √(-135) / 8 = 0. The only real solution to this equation is when the √(-135) term is equal to 0, so we can simplify the expression by setting it equal to 0 and solving for x. We get: √(-135) = 0 or -135 = 0. Since a negative number cannot be equal to zero, only the first solution is valid. This gives us x = 3 as one of our solutions.

Substituting x = 3 into our original equation, 4×2 – 3x + 9 = 2x + 1, we get 4(3)2 – 3(3) + 9 = 2(3) + 1 or 36 – 9 + 9 = 6 + 1, which simplifies to 36 = 7, which is true. Therefore, x = 3 is one of the solutions to our original equation.

To find the other solution for x, we can use the quadratic formula again with a slightly different value for c (9 – 1 = 8). Plugging these values into our quadratic formula, we get: x = -(-3) ± √((-3)2 – 4(4)(8)) / 2(4). Simplifying this expression, we get: x = 3 ± √(9 – 32) / 8. Since we are looking for the values of x that make our equation true, we need to find the two values of x that make our simplified expression equal to 0. To do this, we can set our equation equal to 0 and solve for x. We get: 3 ± √(9 – 32) / 8 = 0. Simplifying this equation, we get: 3 ± √(-23) / 8 = 0. The only real solution to this equation is when the √(-23) term is equal to 0, so we can simplify the expression by setting it equal to 0 and solving for x. We get: √(-23) = 0 or -23 = 0. Since a negative number cannot be equal to zero, only the first solution is valid. This gives us x = 3 as one of our solutions.

Substituting x = -3 into our original equation, 4×2 – 3x + 9 = 2x + 1, we get 4(-3)2 – 3(-3) + 9 = 2(-3) + 1 or -36 + 9 + 9 = -6 + 1, which simplifies to -36 = -5, which is true. Therefore, x = -3 is the other solution to our original equation.

The two solutions for x in the equation 4×2 – 3x + 9 = 2x +1 are x = 3 and x = -3.

## The Values of X

There are two values of x that satisfy the equationx^2 – x + = x + . These values are x = and x = .

To see why these values work, let’s plug them back into the equation:

x = :

( )^2 – ( ) + = ( ) +
= – +
=
which is true!

x = :
( )^2 – ( ) + = ( ) +
= 0
which is also true.

## Conclusion

In conclusion, solving this equation using the quadratic formula can yield two solutions: x = 1 and x = 4. Knowing the values of x is essential to understanding what kind of linear equation this is; in this case, it is a second-degree polynomial with two distinct factors. By using the quadratic formula and some basic algebraic knowledge, we have been able to successfully solve for both factors of our equation and understand its structure more fully.

2. Have you ever been in a situation where you need to solve a quadratic equation but you don’t know where to begin? Don’t worry, you are not alone! The quadratic equation is a type of equation that can take a bit of practice to solve, but thankfully, there is a method that can help you solve it quickly and easily!

The quadratic equation can be solved by using the Quadratic Formula. The formula states that if the equation is in the form ax² + bx + c = 0, then the solution can be found by the following formula: x = [-b ± √(b² – 4ac)] / 2a.

Now, let’s use this formula to solve our example equation: 4x² – 3x + 9 = 2x + 1. In this equation, a = 4, b = -3, and c = 9. Plugging these values into the formula, we get: x = [-(-3) ± √((-3)² – 4(4)(9))] / 2(4), which simplifies to x = [-(-3) ± √(9 – 144)] / 8.

After simplifying further, we get x = [3 ± √(-135)] / 8. Since the square root of a negative number is not a real number, the equation has no real solution, and the values of x are both imaginary numbers.

There you have it! With the help of the Quadratic Formula, we were able to solve our equation and determine that the values of x are imaginary numbers. Now that you know how to use the Quadratic Formula, you can start solving any quadratic equation with ease!