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How do you solve $12{{x}^{2}}-7x+1=0$?

Answer
VerifiedVerified
564.6k+ views
Hint: We have a quadratic equation. We will compare the given quadratic equation with the standard quadratic equation. Then, we will obtain the values of the corresponding coefficients. We will use the quadratic formula to obtain the solution of the given quadratic equation. We will substitute the corresponding coefficients in the formula and solve it to get the solution.

Complete step by step answer:
The given quadratic equation is $12{{x}^{2}}-7x+1=0$. The standard quadratic equation is given as $a{{x}^{2}}+bx+c=0$ where $a$, $b$ and $c$ are the coefficients and $x$ is the variable. Now, we will compare the given quadratic equation with the standard quadratic equation. After comparing the corresponding coefficients of both the quadratic equations, we obtain $a=12$, $b=-7$ and $c=1$.
We will use the quadratic formula to obtain the solution of the given quadratic equation. The quadratic formula is given as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now, we will substitute the values $a=12$, $b=-7$ and $c=1$ in the above formula. We get the following,
$x=\dfrac{-\left( -7 \right)\pm \sqrt{{{\left( -7 \right)}^{2}}-4\times 12\times 1}}{2\times 12}$
Simplifying the above equation, we get
$x=\dfrac{7\pm \sqrt{49-48}}{24}$
Further simplification gives us the following,
$\begin{align}
  & x=\dfrac{7\pm \sqrt{1}}{24} \\
 & \therefore x=\dfrac{7\pm 1}{24} \\
\end{align}$

Hence, we have $x=\dfrac{7+1}{24}$ and $x=\dfrac{7-1}{24}$. This means that $x=\dfrac{8}{24}$ and $x=\dfrac{6}{24}$. We can further reduce these fractions to obtain $x=\dfrac{1}{3}$ and $x=\dfrac{1}{4}$, which are the solutions to the given quadratic equation.

Note: There are different methods to solve quadratic equations. These are the factorization method, the completing square method and the quadratic formula method. We can choose either one of these methods to solve the given equation. The choice depends on the convenience and ease in calculations. Explicitly doing the calculations helps us avoid minor mistakes and obtain the correct solution.