Question

How do you solve $\dfrac{4}{{{x}^{2}}+3x-10}-\dfrac{1}{{{x}^{2}}-x-6}=\dfrac{3}{{{x}^{2}}-x-12}$?

Answer 1

Hint:Here in this question, the given expression $\dfrac{4}{{{x}^{2}}+3x-10}-\dfrac{1}{{{x}^{2}}-x-6}=\dfrac{3}{{{x}^{2}}-x-12}$ consists of quadratic equations. In algebra, a quadratic equation is any equation which can be written in the standard form as $a{{x}^{2}}+bx+c=0$ where $x$ represent an unknown variable, and $a,b,c$ are constants, where $a\ne 0$.

Complete step-by-step solution:
Now considering from the question, the given expression is:
$\dfrac{4}{{{x}^{2}}+3x-10}-\dfrac{1}{{{x}^{2}}-x-6}=\dfrac{3}{{{x}^{2}}-x-12}$.
Now we will solve all the denominators by using the factorization method.
For the first one ${{x}^{2}}+3x-10$ the simplification will be as follows:
$\begin{align}
  & \Rightarrow {{x}^{2}}+5x-2x-10 \\
 & \Rightarrow \left( x+5 \right)\left( x-2 \right) \\
\end{align}$ .
For the next one ${{x}^{2}}-x-6$ the simplification will be as follows:
$\begin{align}
  & \Rightarrow {{x}^{2}}-3x+2x-6 \\
 & \Rightarrow \left( x-3 \right)\left( x+2 \right) \\
\end{align}$ .
For the last one ${{x}^{2}}-x-12$ the simplification will be as follows:
$\begin{align}
  & \Rightarrow {{x}^{2}}-4x+3x-12 \\
 & \Rightarrow \left( x-4 \right)\left( x+3 \right) \\
\end{align}$ .
Now we will use these simplified expressions in the given expression. By doing that we will have
$\Rightarrow \dfrac{4}{\left( x+5 \right)\left( x-2 \right)}-\dfrac{1}{\left( x-3 \right)\left( x+2 \right)}=\dfrac{3}{\left( x-4 \right)\left( x+3 \right)}$ .
Now we will simplify the expressions then we will have $ \dfrac{4}{\left( x+5 \right)\left( x-2 \right)}-\dfrac{1}{\left( x-3 \right)\left( x+2 \right)}=\dfrac{3}{\left( x-4 \right)\left( x+3 \right)}$.
We can easily see that there is no common factor in the above the expression we have, so we leave it as it is and simplify it, now we can rewrite the equation as:
$\Rightarrow \dfrac{4}{{{x}^{2}}+3x-10}-\dfrac{1}{{{x}^{2}}-x-6}-\dfrac{3}{{{x}^{2}}-x-12}=0$ .
Now we will take LCM of the above equation$\dfrac{4}{{{x}^{2}}+3x-10}-\dfrac{1}{{{x}^{2}}-x-6}-\dfrac{3}{{{x}^{2}}-x-12}=0$, and then we will simplify the whole expression into simple form so that we can easily find out the roots of the given expression, then we get
$\begin{align}
  & \Rightarrow \dfrac{4\left( {{x}^{2}}-x-6 \right)\left( {{x}^{2}}-x-12 \right)-\left( {{x}^{2}}+3x-10 \right)\left( {{x}^{2}}-x-12 \right)-3\left( {{x}^{2}}+3x-10 \right)\left( {{x}^{2}}-x-6 \right)}{\left( {{x}^{2}}-x-6 \right)\left( {{x}^{2}}-x-12 \right)\left( {{x}^{2}}+3x-10 \right)} \\
 & \Rightarrow \dfrac{4\left( {{x}^{4}}-2{{x}^{3}}-17{{x}^{2}}+18x+72 \right)-\left( {{x}^{4}}+2{{x}^{3}}-25{{x}^{2}}-26x+120 \right)-3\left( {{x}^{4}}+2{{x}^{3}}-19{{x}^{2}}-8x+60 \right)}{\left( {{x}^{2}}-x-6 \right)\left( {{x}^{2}}-x-12 \right)\left( {{x}^{2}}+3x-10 \right)} \\
 & \Rightarrow \dfrac{4{{x}^{4}}-8{{x}^{3}}-68{{x}^{2}}+72x+288-{{x}^{4}}-2{{x}^{3}}+25{{x}^{2}}+26x-120-3{{x}^{4}}-6{{x}^{3}}+57{{x}^{2}}+24x-180}{\left( {{x}^{2}}-x-6 \right)\left( {{x}^{2}}-x-12 \right)\left( {{x}^{2}}+3x-10 \right)} \\
\end{align}$.
Now we will subtract the same terms in the above equation then we get
$\begin{align}
  & \Rightarrow \dfrac{-16{{x}^{3}}+14{{x}^{2}}+122x-12}{\left( {{x}^{2}}-x-6 \right)\left( {{x}^{2}}-x-12 \right)\left( {{x}^{2}}+3x-10 \right)} \\
 & \Rightarrow \dfrac{-2\left( 8{{x}^{3}}-7{{x}^{2}}-61x+6 \right)}{\left( {{x}^{2}}-x-6 \right)\left( {{x}^{2}}-x-12 \right)\left( {{x}^{2}}+3x-10 \right)} \\
\end{align}$.
So we get the solution of the given equation are the roots of:
$\Rightarrow 8{{x}^{3}}-7{{x}^{2}}-61x+6=0$ .
Hence we can conclude that the simplified form of the expression is $8{{x}^{3}}-7{{x}^{2}}-61x+6=0$ .
So, we have solved the given expression.
Note: In solving these types of questions we can go wrong in the calculation part. The solutions of these types are lengthy and quite complicated to solve. So solve these types of questions carefully. This is the question of the algebra part of mathematics.