Question

How do you solve $x^2+2x=0$?

Answer 1

Hint:
The given quadratic equation $x^2+2x=0$ can be simplified by taking the variable $x$ from both terms. Now the equation will be in the form of the product of two terms and the RHS is equal to zero. You can now put each of the products equal to $0$ and find the value of ‘x’ for which the equation is satisfied. These values will be the roots and solution to the problem.

Complete step by step solution:
Here in this problem, we are given an equation $x^2+2x=0$ . And we need to solve this quadratic equation and find the values of variable ‘x’ that will satisfy the equality.
Before starting with the solution, we must understand a few concepts related to Quadratic equations. In algebra, a quadratic equation is an equation that can be rearranged in standard form as $a{x}^2+bx+c=0$ , where ‘x’ represents an unknown, and ‘a’, ‘b’, and ‘c’ represent known numbers, where $a\ne 0$ . If $a=0$ , then the equation is linear, not quadratic, as there is no term. The numbers ‘a’, ‘b’, and ‘c’ are the coefficients of the equation and may be distinguished by calling them, respectively, the quadratic coefficient, the linear coefficient, and the constant or free term.
The solution of every Quadratic equation will have two possible values called the roots of the equation.
Now let’s try to find the solution to the given equation
$\Rightarrow x^2+2x=0$
There are only two terms in the equation, so we can take common the variable ‘x’
$\Rightarrow x^2+2x=0\Rightarrow x\left(x+2\right)=0$
So now we get the product of two numbers $x$ and $\left(x+2\right)$ as $0$ , which means either one of them can be equal to $0$
$\Rightarrow x=0\text{ or }\left(x+2\right)=0$
This can be further simplified to find the value of $x$ :
$\Rightarrow x=0\text{ or }x=-2$
Therefore, the solution of the equation $x^2+2x=0$ will be $x=0,-2$
We can cross-check our solution by substituting it back to the equation.
For $x=0$ , $x^2+2x=0+2\times 0=0$ , so it satisfies the equation
For $x=-2$ , $x^2+2x={\left(-2\right)}^2+2\times \left(-2\right)=4-4=0$, which also satisfies the equation.

Thus, we get $x=0,-2$ on solving the equation $x^2+2x=0$

Note:
In this question, the method used can only work when we only have terms with variable $x$ and no constant term ‘c’. An alternative approach can be to use the Quadratic formula to solve the given quadratic equation. According to this formula, a quadratic equation of form $a{x}^2+bx+c=0$ has its roots given by $x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ . Notice that here the value of the constant ‘c’ is $c=0$.