Question

How do you solve $2{{x}^{2}}-6x-36=0$?

Answer 1

Hint: We can see that in the equation given in the above question, which is written as $2{{x}^{2}}-6x-36=0$, the factor $2$ is common. So we can divide the given equation by $2$ so as to simplify it as ${{x}^{2}}-3x-18=0$. Then for solving the equation, we need to use the quadratic formula. The quadratic formula is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, where $a$, $b$ and $c$ are respectively the coefficient of ${{x}^{2}}$, the coefficient of $x$, and the constant term. From the equation ${{x}^{2}}-3x-18=0$, the coefficients are $a=1$, $b=-3$ and $c=-18$. On substituting them in the quadratic formula and on simplifying, we will finally get the solutions.

Complete step by step solution:
The equation is
$\Rightarrow 2{{x}^{2}}-6x-36=0$
Since the factor $2$ is common, so we can simplify the above equation by dividing it by $2$ to get
$\Rightarrow {{x}^{2}}-3x-18=0.........\left( i \right)$
Now, we know that the quadratic formula is given by
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
From the equation (i) we can note that $a=1$, $b=-3$ and $c=-18$. Substituting these above, we get
$\begin{align}
  & \Rightarrow x=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\left( 1 \right)\left( -18 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{3\pm \sqrt{9+72}}{2} \\
 & \Rightarrow x=\dfrac{3\pm \sqrt{81}}{2} \\
 & \Rightarrow x=\dfrac{3\pm 9}{2} \\
 & \Rightarrow x=\dfrac{12}{2},x=\dfrac{-6}{2} \\
 & \Rightarrow x=6,x=-3 \\
\end{align}$

Hence, the given equation is solved and the solutions are obtained as $x=6$ and $x=-3$.

Note: We can directly substitute the values of the coefficients from the equation $2{{x}^{2}}-6x-36=0$. It is not necessary to cancel out the common factor of $2$. But the more simplified our quadratic equation is, the less hectic calculations we have to perform after substituting the coefficients into the quadratic formula. So always prefer to do the simplification of the given quadratic equation before proceeding with the solution.