Question

How do you solve $3{{x}^{2}}+19x-14=0$?

Answer 1

Hint: For this problem we need to calculate the solution of the given equation. For a quadratic equation we have so many methods like factorization, completing squares, and quadratic formulas to solve the equation. We can observe that the above equation has large coefficients and constants. So, we are going to use the quadratic formula method to solve this equation. For this we will first compare the given equation with the standard form of the quadratic equation which is $a{{x}^{2}}+bx+c=0$ and write the values as $a=3$, $b=19$, $c=-14$. Now we will substitute those values in the quadratic formula which is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and simplify the equation to get the required solution.

Complete step by step answer:
Given equation $3{{x}^{2}}+19x-14=0$.
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
$a=3$, $b=19$, $c=-14$.
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
$\Rightarrow x=\dfrac{-\left( 19 \right)\pm \sqrt{{{\left( 19 \right)}^{2}}-4\left( 3 \right)\left( -14 \right)}}{2\left( 3 \right)}$
We know that when we multiplied a negative sign with a positive sign, then we will get a negative sign. Applying the above rule and simplifying the above equation, then we will get
$\begin{align}
  & \Rightarrow x=\dfrac{-19\pm \sqrt{361+168}}{6} \\
 & \Rightarrow x=\dfrac{-19\pm \sqrt{529}}{6} \\
\end{align}$
In the above equation we have the value $\sqrt{529}$. We can write the number $529$ as $23\times 23$. Now the value of $\sqrt{529}$ will be $\sqrt{529}=\sqrt{{{23}^{2}}}=23$. Substituting this value in the above equation, then we will get
$\Rightarrow x=\dfrac{-19\pm 23}{6}$
From the above equation, we can write
$\begin{align}
  & \Rightarrow x=\dfrac{-19+23}{6}\text{ or }\dfrac{-19-23}{6} \\
 & \Rightarrow x=\dfrac{4}{6}\text{ or }\dfrac{-42}{6} \\
 & \Rightarrow x=\dfrac{2}{3}\text{ or }-7 \\
\end{align}$

Hence the solution of the given quadratic equation $3{{x}^{2}}+19x-14=0$ are $-7$, $\dfrac{2}{3}$.

Note: In this problem they don’t have mentioned any method to solve the equation. Then we are free to use any of the methods from factoring, quadratic formula or completing squares. Out of all the methods, the quadratic formula is very simple to use for the quadratic equations which are having large numbers because it is very time consuming when you need to multiply them and calculate the factors and such all things. So a quadratic formula is a simple formula without any huge calculations.