Question

How do you solve $3{{x}^{2}}-2x-1=0$?

Answer 1

Hint: In this problem we need to solve the given quadratic equation i.e., we need to calculate the values of $x$ where the given equation is satisfied. For solving a quadratic equation, we have several methods. But in the problem, we are using the quadratic formula which is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Now we will compare the given equation with the standard quadratic equation $a{{x}^{2}}+bx+c=0$ and write the values of $a$, $b$, $c$. Now we will substitute those values in the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and simplify the obtained equation to get the required result.

Complete step by step answer:
Given equation $3{{x}^{2}}-2x-1=0$.
Comparing the above quadratic equation with standard quadratic equation $a{{x}^{2}}+bx+c=0$, then we will get the values of $a$, $b$, $c$ as
$a=3$, $b=-2$, $c=-1$.
We have the quadratic formula for the solution as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the values of $a$, $b$, $c$ in the above equation, then we will get
$\Rightarrow x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 3 \right)\left( -1 \right)}}{2\left( 3 \right)}$
We know that when we multiplied a negative sign with the negative sign, then we will get positive sign. Applying the above rule and simplifying the above equation, then we will get
$\begin{align}
  & \Rightarrow x=\dfrac{2\pm \sqrt{4+12}}{6} \\
 & \Rightarrow x=\dfrac{2\pm \sqrt{16}}{6} \\
\end{align}$
In the above equation we have the value $\sqrt{16}$. We can write the number $16$ as $4\times 4$. Now the value of $\sqrt{16}$ will be $\sqrt{16}=\sqrt{{{4}^{2}}}=4$. Substituting this value in the above equation, then we will get
$\Rightarrow x=\dfrac{2\pm 4}{6}$
From the above equation, we can write
$\begin{align}
  & \Rightarrow x=\dfrac{2+4}{6}\text{ or }\dfrac{2-4}{6} \\
 & \Rightarrow x=\dfrac{6}{6}\text{ or }\dfrac{-2}{6} \\
 & \Rightarrow x=1\text{ or }\dfrac{-1}{3} \\
\end{align}$

Hence the solution of the given quadratic equation $3{{x}^{2}}-2x-1=0$ are $1$, $-\dfrac{1}{3}$.

Note: We can also solve this using factoring or completing square methods. Which are alternatives to this problem. We know that roots are values on the x axis where y is equal to zero.