Question

How do you factor and solve $x(2x - 1) = 0$ ?

Answer 1

Hint:In order to determine the solution to the above equation, since this equation is already in its factor form so no further factorisation is possible. First solution to the equation can be obtained by dividing the both sides of the equation with $(x)$, and similarly to determine the second solution divide both sides with $(2x - 1)$, you will get your required solution.

Complete step by step solution:
We are given a quadratic expression having one variable $x$ i.e. $x(2x - 1) = 0$
$ \Rightarrow x(2x - 1) = 0$ ---------(1)
As we can see the expression is already in factor form and no more factorisation can be possible.
We will move towards the next step of finding the solution.
Since this is a quadratic form, so it will have to have two solutions.
Now, Finding the first solution, by dividing both sides of the equation (1) with $(x)$, we get
$
\dfrac{1}{x}\left( x \right)(2x - 1) = 0 \times \dfrac{1}{x} \\
2x - 1 = 0 \\
$
Transposing $1$ on right hand side of the equation
$2x = 1$
Dividing both sides by the coefficient of variable $x$i.e. 2.
$
\dfrac{{2x}}{2} = \dfrac{1}{2} \\
x = \dfrac{1}{2} \\
$
So our first solution is $x = \dfrac{1}{2}$.
Now for the second solution we will be dividing the equation (1) with $(2x - 1)$, we get
$
\dfrac{1}{{(2x - 1)}}\left( x \right)(2x - 1) = 0 \times \dfrac{1}{{(2x - 1)}} \\
x = 0 \\
$
We get second solution at $x = 0$
Therefore, the solution to the equation$x(2x - 1) = 0$ is $x = 0$ and $x = \dfrac{1}{2}$.
Additional Information:
Quadratic Equation: A quadratic equation is an equation which can be represented in the form of $a{x^2} + bx + c$ where $x$ is the unknown variable and a,b,c are the numbers known where $a \ne 0$. If $a = 0$ then the equation will become a linear equation and will no longer be quadratic . The degree of the quadratic equation is of the order 2. Every Quadratic equation has 2 roots.

Note:
1. One must be careful while calculating the answer as calculation error may come.
2.Every equation in one variable having degree of order 2 will always have two solutions for the equation.