X 2 2x 5 0

Solving the Quadratic Equation ( x^2 + 2x - 5 = 0 ): A Complete Guide

At first glance, the equation ( x^2 + 2x - 5 = 0 ) appears as a simple string of symbols. Yet, within its structure lies a gateway to fundamental algebraic principles and problem-solving techniques used across science, engineering, and economics. This quadratic equation—a polynomial of degree two—serves as an perfect case study for understanding how to find unknown values that satisfy a mathematical relationship. Unlike simpler linear equations, quadratics can have two distinct solutions, one repeated solution, or even complex solutions involving imaginary numbers. The specific equation ( x^2 + 2x - 5 = 0 ) is particularly instructive because it does not factor neatly with integers, forcing us to employ more powerful and general methods. Mastering its solution builds a robust foundation for tackling countless real-world problems, from calculating the trajectory of a projectile to optimizing business profits.

Detailed Explanation: What is a Quadratic Equation?

A quadratic equation is any equation that can be rearranged into the standard form ( ax^2 + bx + c = 0 ), where ( a ), ( b ), and ( c ) are known numbers (called coefficients), ( a \neq 0 ), and ( x ) represents the unknown variable. The highest power of the unknown is two, which is the defining characteristic. The solutions to this equation are called roots or zeros, as they correspond to the x-values where the graph of the quadratic function ( y = ax^2 + bx + c ) crosses the x-axis.

Our equation, ( x^2 + 2x - 5 = 0 ), fits this form perfectly:

• ( a = 1 ) (the coefficient of ( x^2 ))
• ( b = 2 ) (the coefficient of ( x ))
• ( c = -5 ) (the constant term)

The goal is to find the value(s) of ( x ) that make the left-hand side equal to zero. The presence of the ( x^2 ) term means the relationship is non-linear, and its graph is a parabola. For this specific equation, since ( a = 1 > 0 ), the parabola opens upwards. The fact that ( c = -5 ) tells us the y-intercept is at (0, -5), below the x-axis. This visual hint suggests the parabola will cross the x-axis at two points, implying two real solutions. We need algebraic tools to find their exact values.

Step-by-Step Breakdown: The Quadratic Formula

When a quadratic equation does not factor easily into two binomials with integer coefficients—as is the case with ( x^2 + 2x - 5 = 0 )—the most reliable method is the quadratic formula. This formula is derived from the process of completing the square and provides a direct solution for any quadratic equation in standard form.

The quadratic formula is: [ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]

The expression under the square root, ( b^2 - 4ac ), is called the discriminant. It is the key to predicting the nature of the solutions before we even calculate them:

• If ( b^2 - 4ac > 0 ), there are two distinct real solutions.
• If ( b^2 - 4ac = 0 ), there is exactly one real solution (a repeated root).
• If ( b^2 - 4ac < 0 ), there are two complex (non-real) solutions.

Let's apply this formula step-by-step to our equation ( x^2 + 2x - 5 = 0 ):

1. Identify Coefficients: Clearly state ( a = 1 ), ( b = 2 ), ( c = -5 ). Writing these down prevents sign errors, which are the most common mistake.
2. Calculate the Discriminant: ( b^2 - 4ac = (2)^2 - 4(1)(-5) = 4 - (-20) = 4 + 20 = 24 ). Since 24 is positive and not a perfect square, we know we will have two distinct, irrational real solutions.
3. Substitute into the Formula: [ x = \frac{-(2) \pm \sqrt{24}}{2(1)} = \frac{-2 \pm \sqrt{24}}{2} ]
4. Simplify the Radical: ( \sqrt{24} ) can be simplified. ( 24 = 4 \times 6 ), and ( \sqrt{4} = 2 ), so ( \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6} ).
5. Final Simplification: [ x = \frac{-2 \pm 2\sqrt{6}}{2} ] We can factor a 2 out of the numerator: [ x = \frac{2(-1 \pm \sqrt{6})}{2} ] Canceling the 2 gives the final, simplified exact solutions: [ x = -1 \pm \sqrt{6} ] Therefore, the two solutions are: [ x_1 = -1 + \sqrt{6} \quad \text{and} \quad x_2 = -1 - \

Therefore, the two solutions are: [ x_1 = -1 + \sqrt{6} \quad \text{and} \quad x_2 = -1 - \sqrt{6} ] In decimal form, these are approximately ( x_1 \approx 1.449 ) and ( x_2 \approx -3.449 ), confirming the parabola crosses the x-axis at two distinct points as predicted by the positive