Question

How do you solve \[2{x^2} - 2x - 12 = 0\] ?

Answer 1

Hint: The given equation is a quadratic equation and its roots can be calculated by the use of sridharacharya formula.
Sridharacharya formula: If a quadratic is given by $a{x^2} + bx + c = 0$ then its roots can be given by $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ where $D$ is the discriminant of the given quadratic equation which is equal to ${b^2} - 4ac$.

Complete step-by-step answer:
The given quadratic equation is \[2{x^2} - 2x - 12 = 0\].
Taking $2$ as common from each term we can write this quadratic equation as $2\left( {{x^2} - x - 6} \right) = 0$
Now, we have to find the roots of quadratic equation ${x^2} - x - 6 = 0$
Here, the coefficient of ${x^2}$ is $a = 1$, the coefficient of $x$ is $b = - 1$ and the constant $c = - 6$.
Now, by using the above given Shridharacharya formula we can write the roots $x = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4 \times 1 \times \left( { - 6} \right)} }}{{2 \times 1}}$.
Or, $x = \dfrac{{1 \pm \sqrt {1 + 24} }}{2}$
Or, $x = \dfrac{{1 \pm 5}}{2}$
Thus, ${x_1} = \dfrac{{1 + 5}}{2} = \dfrac{6}{2} = 3$ and ${x_2} = \dfrac{{1 - 5}}{2} = \dfrac{{ - 4}}{2} = - 2$
Thus, the roots of the given quadratic equation are $ - 2$ and $3$.

Hence, $ - 2$ and $3$ are the solution of the given equation \[2{x^2} - 2x - 12 = 0\].

Note:
The given quadratic equation ${x^2} - x - 6 = 0$ can be solved by factorization method. For solving quadratic equations by factor method, we have to break the term of $x$ in two parts such that their product is equal to the product of constant term and the term having ${x^2}$ that is $ - 6{x^2}$.
The quadratic equation ${x^2} - x - 6 = 0$ can be written as ${x^2} - 3x + 2x - 6 = 0$.
Now taking $x$ as common from the first two terms and $2$ from the last two terms. We can write,
Or, $x\left( {x - 3} \right) + 2\left( {x - 3} \right) = 0$
Then taking the term $\left( {x - 3} \right)$ common. We can write
Or, $\left( {x - 3} \right)\left( {x + 2} \right) = 0$
Or, $x - 3 = 0$ and $x + 2 = 0$
Thus, $x = - 2$ and $x = 3$are the solution of a given quadratic equation.
Alternatively:
This is a quadratic equation so, let $\alpha $ and $\beta $ are its two roots.
We can write the sum of roots $\alpha + \beta = \dfrac{{ - b}}{a} = \dfrac{{ - \left( { - 1} \right)}}{1} = 1$ --------(1)
And the product of roots $\alpha \beta = \dfrac{c}{a} = \dfrac{{ - 6}}{1} = - 6$------(2)
By solving equation (1) and (2) we get both the roots of the given quadratic equation.