# For what values of n, will the equation x^2=n have two solutions?

## Question:

For what values of n, will the equation {eq}x^2=n {/eq} have two solutions?

## Solving Quadratic Equation: Two Solutions

A quadratic equation in its standard form would be written as {eq}ax^2 + bx + c = 0 {/eq}. The coefficients {eq}a {/eq}, {eq}b {/eq} and {eq}c {/eq} indirectly define both the number and the nature of the solutions. Specifically, a discriminant value defined by {eq}D = b^2 - 4ac {/eq} tells us whether solutions will be real or imaginary, as well as whether there will be one or two distinct solutions.

Given an equation:

{eq}x^2 = n {/eq}

Rearrange it into a quadratic form:

{eq}x^2 - n = 0 {/eq}

Let's identify the coefficients:

{eq}a = 1\\ b = 0\\ c = -n {/eq}

Using the discriminant formula:

{eq}D = b^2 - 4ac = 0^2 - 4\cdot 1\cdot (-n) = 4n {/eq}

We know that if the value of {eq}D {/eq} is equal to 0, then there will be one distinct solution with a multiplicity of 2. Therefore, if we wish to have two distinct solutions:

{eq}D\neq 0\\ \therefore 4n\neq 0\\ n\neq 0 {/eq}

Thus, for any value {eq}n\neq 0 {/eq}, we will have either two real or two imaginary solutions.

Alternatively, we can avoid using discriminant and take square root of both sides:

{eq}x^2 = n\\ \sqrt{x^2} = \sqrt{n}\\ |x| = \sqrt{n}\\ x = \pm \sqrt{n} {/eq}

A positive and a negative number will only be equal to each other if that number is 0, thus, we can conclude that the given equation will have two solutions for any {eq}\boxed{n\neq 0} {/eq}