Question

How do you solve $3{{x}^{2}}-14x-5=0$?

Answer 1

Hint: For this problem we need to solve the given equation which is a quadratic equation. We can observe that the given equation is in the form of $a{{x}^{2}}+bx+c=0$. We can use the quadratic formula, completing the square method, factoring the quadratic equation to solve the given equation. But for now, we are sticking with the factorization method which uses the factors of the given equation to find the solution. So, we need to first calculate the factors of the given equation. For this we will compare the given equation with $a{{x}^{2}}+bx+c=0$ and write the values of $a$, $b$, $c$. Now we will split the term $bx$ as ${{b}_{1}}x+{{b}_{2}}x$ where ${{b}_{1}}{{b}_{2}}=ac$. So, we will calculate the value of $ac$ and from the factors of the $ac$ we will split the term. After that we will take appropriate terms as common from the equation and equate it to zero to get the solution.

Complete step by step answer:
Given the equation, $3{{x}^{2}}-14x-5=0$.
Comparing the above equation with $a{{x}^{2}}+bx+c=0$. Then we will get
$a=3$, $b=-14$, $c=-5$.
Now the value of $ac$ will be $ac=3\left( -5 \right)=-15$. We have the factors of $15$ as $1$, $3$, $5$, $15$. From these factors we can write
$\begin{align}
  & -15+1=-14 \\
 & -15\times 1=-15 \\
\end{align}$
Now splitting the middle term as $-14x=15x+x$. Now the above equation is modified as
$\Rightarrow 3{{x}^{2}}-15x+x-5=0$
Taking $3x$ as common from the first two terms, then we will get
$\Rightarrow 3x\left( x-5 \right)+1\left( x-5 \right)=0$
Now taking $x-5$ as common from the above equation, then we will get
$\Rightarrow \left( x-5 \right)\left( 3x+1 \right)=0$
Now equating both the factors to zero, then we will have
$x-5=0$ or $3x+1=0$.
Considering the equation $x-5=0$. From this equation we can write
$x=5$.
Considering the equation $3x+1=0$. From this equation we can write
$\begin{align}
  & \Rightarrow 3x+1=0 \\
 & \Rightarrow 3x=-1 \\
 & \Rightarrow x=-\dfrac{1}{3} \\
\end{align}$

Hence the roots of the given equation $3{{x}^{2}}-14x-5=0$ are $x=5,-\dfrac{1}{3}$.

Note: We can also check whether the obtained solution is correct or not. We will substitute the obtained roots in the given equation and check whether the given equation is satisfied or not.
Substituting $x=5$ in the given equation $3{{x}^{2}}-14x-5=0$, then we will get
$\begin{align}
  & \Rightarrow 3{{\left( 5 \right)}^{2}}-14\left( 5 \right)-5=0 \\
 & \Rightarrow 3\left( 25 \right)-70-5=0 \\
 & \Rightarrow 75-75=0 \\
 & \Rightarrow 0=0 \\
 & \Rightarrow LHS=RHS \\
\end{align}$
Hence the obtained solution is correct.