Question

How do you solve: $6{x^2} + x = 2$ ?

Answer 1

Hint: Given equation is of degree $2$. Equations of degree $2$ are known as quadratic equations. Quadratic equations can be factored by the help of splitting the middle term method. In this method, the middle term is split into two terms in such a way that the equation remains unchanged.

Complete step-by-step answer:
For solving the given quadratic equation $6{x^2} + x = 2$ , we will use the splitting the middle term method.
First transpose the constant term from the right side of the equation to the left side of the equation.
So, $6{x^2} + x = 2$
$ \Rightarrow 6{x^2} + x - 2 = 0$
Now, we have to factorise the quadratic equation thus obtained. We can use the splitting of the middle term method in which the middle term is split into two terms such that the sum of the terms gives us the original middle term and the product of the terms gives us the product of the constant term and coefficient of ${x^2}$.
$ \Rightarrow 6{x^2} + x - 2 = 0$
We split the middle term $x$ into two terms $4x$ and $ - 3x$ since the product of these terms, $ - 12{x^2}$ is equal to the product of the constant term and coefficient of ${x^2}$ and sum of these terms gives us the original middle term, $x$.
$ \Rightarrow 6{x^2} + 4x - 3x - 2 = 0$
Taking $2x$ common from the first two terms and negative sign common from the last two terms. We get,
$ \Rightarrow 2x\left( {3x + 2} \right) - \left( {3x + 2} \right) = 0$
Taking the bracket as common, we get,
$ \Rightarrow \left( {2x - 1} \right)\left( {3x + 2} \right) = 0$
Now, if the product of two terms is zero, then either of the terms has to be zero.
So, we get,
Either $\left( {2x - 1} \right) = 0$ or $\left( {3x + 2} \right) = 0$
So, we get,
$ \Rightarrow 2x = 1$ or $ \Rightarrow 3x = - 2$
Shifting the terms and finding the values of x, we get,
$ \Rightarrow x = \dfrac{1}{2}$ or $ \Rightarrow x = - \dfrac{2}{3}$
So, the correct answer is “$x = \dfrac{1}{2}$ and $x = - \dfrac{2}{3}$”.

Note: Splitting of middle term can be a tedious process at times when the product of the constant term and coefficient of ${x^2}$ is a large number with a large number of divisors. Special care should be taken in such cases. Similar to quadratic polynomials, quadratic solutions can also be solved using the factorisation method. Prime factorisation of the product of the constant term and coefficient of ${x^2}$ should be done to find the possible pairs that sum up to the middle term of the equation.