Question

How do you solve $3{{x}^{2}}-4x=4$?

Answer 1

Hint: For this problem we need to calculate the solution of the given equation. We can observe that the given equation is a quadratic equation but it is not in standard form. So, we will perform some arithmetic operations to convert the given equation into standard form which is $a{{x}^{2}}+bx+c=0$. After converting it we will compare the obtained equation with the standard form of the equation and write the values $a=3$, $b=-4$, $c=-4$. Now we will use the quadratic formula which is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve the equation. So, we will substitute the values we have in the above equation and simplify it to get the required solution.

Complete step by step answer:
Given equation $3{{x}^{2}}-4x=4$.
Shifting the constant which is right hand side to left hand side, then we will get
$\Rightarrow 3{{x}^{2}}-4x-4=0$
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
$a=3$, $b=-4$, $c=-4$.
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
$\Rightarrow x=\dfrac{-\left( -4 \right)\pm \sqrt{{{\left( -4 \right)}^{2}}-4\left( 3 \right)\left( -4 \right)}}{2\left( 3 \right)}$
We know that when we multiplied a negative sign with the negative sign, then we will get a positive sign. Applying the above rule and simplifying the above equation, then we will get
$\begin{align}
  & \Rightarrow x=\dfrac{4\pm \sqrt{16+48}}{6} \\
 & \Rightarrow x=\dfrac{4\pm \sqrt{64}}{6} \\
\end{align}$
In the above equation we have the value $\sqrt{64}$. We can write the number $64$ as $8\times 8$. Now the value of $\sqrt{64}$ will be $\sqrt{64}=\sqrt{{{8}^{2}}}=8$. Substituting this value in the above equation, then we will get
$\Rightarrow x=\dfrac{4\pm 8}{6}$
Taking $4$ as common and simplifying the above equation, then we will get
$\begin{align}
  & \Rightarrow x=\dfrac{4\left( 1\pm 2 \right)}{6} \\
 & \Rightarrow x=\dfrac{2\pm 4}{3} \\
\end{align}$
From the above equation, we can write
$\begin{align}
  & \Rightarrow x=\dfrac{2+4}{3}\text{ or }\dfrac{2-4}{3} \\
 & \Rightarrow x=\dfrac{6}{3}\text{ or }\dfrac{-2}{3} \\
 & \Rightarrow x=2\text{ or }-\dfrac{2}{3} \\
\end{align}$

Hence the solution of the given quadratic equation $3{{x}^{2}}-4x=4$ are $2$, $-\dfrac{2}{3}$.

Note: There are several methods to solve the quadratic equation. But the easiest one is the using quadratic formula. There are less chances of making mistakes in this method. So, when they don’t mention the method to use, we can freely use this method and solve the equation.