Question

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

Answer 1

We are given$2{{x}^{2}}-x=0$, we are asked to find the factor of this, to do so we will first understand the type of equation we have, once we get that we reduce the given equation into the standard form to simplify after that we will find the greatest common factor from each term then in the remaining term be factor using the middle term and lastly we compare it with zero and solve further, We use arc in such a way that its sum or difference from the ‘b’ of the equation $a{{x}^{2}}+bx+c=0$

Complete step-by-step solution:
We are given $2{{x}^{2}}-x=0x$ we are asked to find the factor of it.
Now to find the factor of the equation, we should see that as the highest power is 2 so it is or 2 degree polynomial. So it is a quadratic equation.
As our Equation is quadratic equations and we know quadratic equations are given as $a{{x}^{2}}+bx+c=0$. We reduce it to the standard form
We can see that our equation$2{{x}^{2}}-x=0$ is already in standard form
Now, to find its factor we will first find the possible greatest common factor of all these.
In 2 and -1 we can see that 1 is the only possible term that can be separated, so we get –
Our equation stays like original a$2{{x}^{2}}-x=0$
We can write this above equation as
$2{{x}^{2}}-x=0$
Now we will use the middle term to split
In middle term split apply on $a{{x}^{2}}+bx+c=0$, we produce ‘a’ by ‘c’ and then factor ‘ac’ in such a way that if the product is ‘ac’ while the sum or difference is made up to ‘b’.
Now, we have middle term split on $2{{x}^{2}}-x=0$
We have a=2 b=-1 and c=0
So,
$\Rightarrow$ $axc=2\times 0=0$
We have to find number whose sum or difference is -1 and product is0 as product is zero one must be zero
Now we can see that $0\times -1=0$ and$0-1=-1$
So we use this to split the middle term. So,
$2{{x}^{2}}-x+0=2{{x}^{2}}+\left( 0-1 \right)x+0$
$2{{x}^{2}}-2x+0= 2{{x}^{2}}+(0-1)x+0$
Opening brackets we get
$2{{x}^{2}}+0x-x+0$
We take common in the first 2 terms and the last 2 terms. So we get –
$=x\left( 2x+0-1+0 \right)$ [We can anything out of 0 as $0=0\times x$ ] As x is same, so we get –
$=x\left( 2x+0-1+0 \right)$
Simplifying we get
$=x\left( 2x-1 \right)$
So, we get –
$\Rightarrow$ $2{{x}^{2}}-x+0=x\left( 2x-1 \right)$
Simplifying we get factor as $2{{x}^{2}} -x=x(2x-1)$
Putting this in our original equation we get
$x\left( 2x-1 \right)=0$
Now using product rule we get either $x=0$ or $2x-1=0$
So simplifying we get
$\Rightarrow$ $x=0$ or $x=\dfrac{1}{2}$
Hence solution are$x=0$ and $x=\dfrac{1}{2}$

Note: While find the middle term using factor of $a\times c$, we need to keep in mind that when the sign of ‘a’ and ‘c’ are same then ‘b’ is obtained by addition only, if the sign of ‘a’ and ‘c’ are different then ‘b’ can be obtained using only subtraction.
So, as we have $a=3$ and $c=-3$ different signs so ‘b’ is obtained as $9-2=7$ using subtraction.
Key point to remember that degree of the equation will also tell us about the number of solution also 2 degree means given equation can have only 2 Factor
We can easily find the factor of this
As we can see that our problem just have two terms so we will try to take the greatest possible common term our and then simply
We have $2{{x}^{2}}-x$
We can see x is common to all
We take x from $2{{x}^{2}}$ it will be left with 2x and we take x from -x it will be left with -1
So $2{{x}^{2}}-x=x\left( 2x-1 \right)$We get two factor
It cannot be factorized more.
This is our factor form
$2{{x}^{2}}-x=x\left( 2x-1 \right)$