Question

How do you solve $2{{x}^{2}}-6x=0$ by factoring?

Answer 1

Hint: In this problem we need to calculate the solution of the given equation by factoring the given equation. We can observe that the given equation is a quadratic equation without any constant. So, we can easily factorize the given equation without splitting the middle term or $x$ term. We have only the ${{x}^{2}}$ and $x$ terms in the above equation, so we will take $x$ along with the appropriate coefficient as a common from the given equation. Now we will have the factors of the given equation. To find the solution we will equate each factor to zero individually, and simplify the equations, then we will get the required solution for the given equation.

Complete step by step answer:
Given equation $2{{x}^{2}}-6x=0$.
Taking $2x$ as common from the above equation, then we will have
$2x\left( x-3 \right)=0$
From the above equation we can say that the factors of the given equation $2{{x}^{2}}-6x=0$ are $2x$, $x-3$. Now equating each term individually to zero to get the result.
Equating the term $2x$ to zero, then we will have
$\begin{align}
  & 2x=0 \\
 & \Rightarrow x=0 \\
\end{align}$
Now equating the term $x-3$ to zero, then we will have
$x-3=0$
Adding $3$ on both sides of the above equation, then we will have
$x-3+3=3$
We know that $+a-a=0$, then we will get
$x=3$.

Hence the solution for the given equation $2{{x}^{2}}-6x=0$ is $x=0,3$.

Note: You can also use the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve the above equation. On comparing the given equation with the $a{{x}^{2}}+bx+c=0$, we have $a=2$, $b=-6$, $c=0$. Substituting these values in the above equation and simplifying the equation, then we will get
$\begin{align}
  & \Rightarrow x=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\left( 2 \right)\left( 0 \right)}}{2\left( 2 \right)} \\
 & \Rightarrow x=\dfrac{6\pm \sqrt{36}}{4} \\
 & \Rightarrow x=\dfrac{6\pm 6}{4} \\
 & \Rightarrow x=\dfrac{6-6}{4}\text{ or }\dfrac{6+6}{4} \\
 & \Rightarrow x=0\text{ or }3 \\
\end{align}$
From both the methods we got the same result.