Question

Solve the equation given by $9{{x}^{2}}-3x-2=0$ for variable x.

Answer 1

Hint: As we need to the value of x which satisfies a quadratic equation so, we will use the formula by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ which is a basic formula that is used to find out the roots of any quadratic equation. This will solve our solution in a direct way.

Complete step-by-step answer:
We will consider the equation $9{{x}^{2}}-3x-2=0...(i)$. As it is a quadratic equation. Therefore we will use the formula of square root to find its roots. The formula is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Now substituting a = 9, b = - 3 and c = - 2. Therefore, we will get
 $\begin{align}
  & x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow x=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\left( 9 \right)\left( -2 \right)}}{2\left( 9 \right)} \\
 & \Rightarrow x=\dfrac{3\pm \sqrt{9+72}}{18} \\
 & \Rightarrow x=\dfrac{3\pm \sqrt{81}}{18} \\
 & \Rightarrow x=\dfrac{3\pm 9}{18} \\
\end{align}$
Now, we get the values of x as $x=\dfrac{3+9}{18}$ and $x=\dfrac{3-9}{18}$. Solving these valules further into its simplest form we will get $x=\dfrac{12}{18}$ and $x=\dfrac{-6}{18}$. Or, the values of $x=\dfrac{2}{3}$ and $x=\dfrac{-1}{3}$.
Hence the required values of x for the equation $9{{x}^{2}}-3x-2=0$ are given by $x=\dfrac{2}{3}$ and $x=\dfrac{-1}{3}$.

Note: We could have used an alternate method for this question which is known as hit and trial method. But in this question the values are in fraction. And finding out fractions for substitution is not possible at all. Moreover, if we check for the values of x we will check some of them as done below.
Alternatively we can solve the equation $9{{x}^{2}}-3x-2=0$ by splitting the middle term - 3x into - 6x + 3x. Therefore we get
$\begin{align}
  & 9{{x}^{2}}-3x-2=9{{x}^{2}}-6x+3x-2 \\
 & \Rightarrow 9{{x}^{2}}-3x-2=3x\left( 3x-2 \right)-1\left( 3x-2 \right) \\
 & \Rightarrow 9{{x}^{2}}-3x-2=\left( 3x-1 \right)\left( 3x-2 \right) \\
\end{align}$
By keeping the equation again equal to 0 therefore we get $\left( 3x-1 \right)\left( 3x-2 \right)=0$ resulting into $x=\dfrac{2}{3}$ and $x=\dfrac{-1}{3}$.