what are the two solutions of 2x2 x2 5x 1

less than a minute read 25-12-2024

what are the two solutions of 2x2 x2 5x 1

Solving the Quadratic Equation: 2x² + 5x + 1 = 0

This article explores how to find the two solutions (roots) of the quadratic equation 2x² + 5x + 1 = 0. We'll use two common methods: the quadratic formula and factoring.

Understanding Quadratic Equations

A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b, and c are constants, and a ≠ 0. Our equation, 2x² + 5x + 1 = 0, fits this form with a = 2, b = 5, and c = 1. Quadratic equations typically have two solutions.

Method 1: Using the Quadratic Formula

The quadratic formula provides a direct way to solve for x:

x = [-b ± √(b² - 4ac)] / 2a

Plugging in the values from our equation (a = 2, b = 5, c = 1), we get:

x = [-5 ± √(5² - 4 * 2 * 1)] / (2 * 2) x = [-5 ± √(25 - 8)] / 4 x = [-5 ± √17] / 4

This gives us two solutions:

x₁ = (-5 + √17) / 4 ≈ -0.219
x₂ = (-5 - √17) / 4 ≈ -2.281

Method 2: Factoring (if possible)

While the quadratic formula always works, some quadratic equations can be solved by factoring. Unfortunately, our equation 2x² + 5x + 1 = 0 doesn't factor easily using integers. Let's illustrate factoring with a simpler example:

Consider the equation x² + 5x + 6 = 0. This factors to (x + 2)(x + 3) = 0. The solutions are x = -2 and x = -3.

Since our equation, 2x² + 5x + 1 = 0, doesn't readily factor with integers, the quadratic formula is the more efficient method in this case.

Checking the Solutions

It's always a good practice to check your solutions by substituting them back into the original equation:

For x₁ ≈ -0.219:

2(-0.219)² + 5(-0.219) + 1 ≈ 0.095 + (-1.095) + 1 ≈ 0 (Close enough, considering rounding)

For x₂ ≈ -2.281:

2(-2.281)² + 5(-2.281) + 1 ≈ 10.438 + (-11.405) + 1 ≈ 0 (Close enough, considering rounding)

Conclusion

The two solutions to the quadratic equation 2x² + 5x + 1 = 0 are approximately x ≈ -0.219 and x ≈ -2.281. These were obtained using the quadratic formula, as factoring wasn't straightforward in this case. Remember to always check your solutions by substituting them back into the original equation.