Question

How do you solve $12{{x}^{2}}-6x=0$?

Answer 1

We first try to take common terms out of the given equation $12{{x}^{2}}-6x=0$. We need to form factorisation from the left side equation $12{{x}^{2}}-6x$. We have one variable $x$ and one constant 6 to take as common. From the multiplication we find the solution for $12{{x}^{2}}-6x=0$.

Complete step-by-step solution:
We need to find the solution of the given equation $12{{x}^{2}}-6x=0$.
First, we try to take a common number or variable out of the terms $12{{x}^{2}}$ and $-6x$.
The only thing that can be taken out is $6x$. We have one variable $x$ and one constant 6 to take as common.
So, $12{{x}^{2}}-6x=6x\left( 2x-1 \right)=0$.
The multiplication of two terms gives 0. This gives that at least one of the terms has to be zero.
We get the values of x as either $6x=0$ or $\left( 2x-1 \right)=0$.
This gives $x=0,\dfrac{1}{2}$.
The given quadratic equation has 2 solutions and they are $x=0,\dfrac{1}{2}$.

Note: The highest power of the variable or the degree of a polynomial decides the number of roots or the solution of that polynomial. Quadratic equations have 2 roots. Cubic polynomials have 3. It can be both real and imaginary roots.
We can also apply the quadratic equation formula to solve the equation $12{{x}^{2}}-6x=0$.
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have $12{{x}^{2}}-6x=0$. The values of a, b, c is $12,-6,0$ respectively.
We put the values and get x as $x=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\times 1\times 0}}{2\times 12}=\dfrac{6\pm \sqrt{{{6}^{2}}}}{24}=\dfrac{6\pm 6}{24}=0,\dfrac{1}{2}$.