Question

Solve equations using factorisation method: $2{x^2} - \dfrac{1}{2}x = 0$ .

Answer 1

Hint: As we know that factorizing is the reverse of expanding brackets, it is an important way of solving equations. We know that the first step of factorizing an expression is to take out any common factors which the terms have. To factorize this equation $2{x^2} - \dfrac{1}{2}x = 0$ ,We will multiply the whole equation by $2$ , so that we can eliminate the denominator and then take the common factor out.

Complete answer:
Here we have been given a linear equation $2{x^2} - \dfrac{1}{2}x = 0$ .
Let us multiply the whole equation with $2$, so we can write it as:
$ \Rightarrow 2\left( {2{x^2} - \dfrac{1}{2}x} \right) = 0$
On multiplying the values we have
 $ \Rightarrow 4{x^2} - x = 0$
We can take the common factor out of the equation and we have:
$ \Rightarrow x(4x - 1) = 0$
Therefore it gives us two values:
We have either $x = 0$ or $4x - 1 = 0$
Now simplify the values:
$ \Rightarrow 4x = 1$
$ \Rightarrow x = \dfrac{1}{4}$
Hence it gives us two values: $x = 0$ or $x = \dfrac{1}{4}$

Note:
We should note that if we were asked to factor the expression ${x^2} + x$ ; since $x$ goes into both terms, we would write $x(x + 1)$. We will also use the identities to solve the factorization of equation where needed.
We should some of the basic algebraic identities which will help us to factorize an algebraic expression easily i.e.
${(a - b)^2} = {a^2} - 2ab + {b^2}$ and ${a^2} - {b^2} = (a + b)(a - b)$.
We should keep in mind while solving these expressions that we use correct identities to factorize the given algebraic expressions and keep checking the negative and positive signs otherwise it will give the wrong answer. The above are some of the standard algebraic identities. This is as far we can go with real coefficients as the remaining quadratic factors all have complex zeroes.