Question

Solve the equation to obtain x
$2{{x}^{2}}+x-6=0$

Answer 1

Hint: In this question, we need to find out the value of x from the given expression. We notice that the highest power of x is 2 and thus it is a quadratic equation. Thus, we can use the common methods to solve quadratic equations to solve this problem. One is the factorization method in which we have to find two numbers such that their sum gives the coefficient of x and their product is equal to the product of the coefficients of ${{x}^{2}}$and the constant term.
Complete step-by-step answer:
The given equation is $2{{x}^{2}}+x-6=0$.
We note that as the highest power of x is 2, the equation is a quadratic equation. To solve it using the factorization method, we need to find two numbers such that their sum is equal to 1 and their product is equal to $2\times -6=-12$. We notice that these conditions will be satisfied by the numbers 4 and -3 because
$4+(-3)=1$
And $4\times (-3)=-12$
Thus, we can rewrite the given equation as
$\begin{align}
  & 2{{x}^{2}}+x-6=0=2{{x}^{2}}+(4-3)x-6=0 \\
 & \Rightarrow 2x(x+2)-3(x+2)=0 \\
 & \Rightarrow (2x-3)\left( x+2 \right)=0 \\
\end{align}$
Thus, we have factorized the equation into two factors. Now, the Left hand side will be equal to zero if and only if atleast one of the factors is equal to zero. Thus, the values of x for which the equation will be satisfied are $2x-3=0\Rightarrow x=\dfrac{3}{2}$ and $x+2=0\Rightarrow x=-2$.
So, the answer is $x=\dfrac{3}{2}$ and $x=-2$.
Note: We have obtained two solutions to this problem. We should understand that a quadratic equation can have two solutions for which the given equation holds and thus both solutions are equally valid. We can solve quadratic equations using the quadratic formula and completing squares as well.