Question

Solve the following equation for \['x'\]
\[7x-\dfrac{\sqrt{3{{x}^{2}}-8x+1}}{x}={{\left( \dfrac{8}{\sqrt{x}}+\sqrt{x} \right)}^{2}}\]

Answer 1

Hint: We solve this problem simply by expanding the terms in such a way that we can get the common term as \[\sqrt{3{{x}^{2}}-8x+1}\] in order to reduce the solution. Here we may get the equation as
\[\Rightarrow \sqrt{3{{x}^{2}}-8x+1}=f\left( x \right)\]
Here, squaring on both sides leads to long and difficult problems. So, we search to get the term \[\sqrt{3{{x}^{2}}-8x+1}\] as common from both sides to reduce the difficulty of the problem.

Complete step-by-step solution:
Let us take the given equation as
\[\Rightarrow 7x-\dfrac{\sqrt{3{{x}^{2}}-8x+1}}{x}={{\left( \dfrac{8}{\sqrt{x}}+\sqrt{x} \right)}^{2}}\]
Now, by applying the LCM to LHS and expanding the square on RHS we get
\[\begin{align}
  & \Rightarrow \dfrac{7{{x}^{2}}-\sqrt{3{{x}^{2}}-8x+1}}{x}=\dfrac{64}{x}+x+16 \\
 & \Rightarrow \dfrac{7{{x}^{2}}-\sqrt{3{{x}^{2}}-8x+1}}{x}=\dfrac{64+{{x}^{2}}+16x}{x} \\
 & \Rightarrow 7{{x}^{2}}-\sqrt{3{{x}^{2}}-8x+1}=64+{{x}^{2}}+16x \\
\end{align}\]
Now, by rearranging the terms in the above equation we get
\[\begin{align}
  & \Rightarrow 6{{x}^{2}}-16x-64=\sqrt{3{{x}^{2}}-8x+1} \\
 & \Rightarrow 2\left( 3{{x}^{2}}-8x-32 \right)=\sqrt{3{{x}^{2}}-8x+1} \\
\end{align}\]
Now, let us get the common term that is \[\sqrt{3{{x}^{2}}-8x+1}\] on both sides we get
\[\begin{align}
  & \Rightarrow 2\left( \left( 3{{x}^{2}}-8x+1 \right)-33 \right)=\sqrt{3{{x}^{2}}-8x+1} \\
 & \Rightarrow 2\left( {{\left( \sqrt{3{{x}^{2}}-8x+1} \right)}^{2}}-33 \right)=\sqrt{3{{x}^{2}}-8x+1}.......equation(i) \\
\end{align}\]
Let us assume that
\[\Rightarrow \sqrt{3{{x}^{2}}-8x+1}=t\]
By substituting the above equation in equation (i) we get
\[\begin{align}
  & \Rightarrow 2\left( {{t}^{2}}-33 \right)=t \\
 & \Rightarrow 2{{t}^{2}}-t-66=0 \\
\end{align}\]
By using the factorisation method to the above equation we get
\[\begin{align}
  & \Rightarrow 2{{t}^{2}}-12t+11t-66=0 \\
 & \Rightarrow 2t\left( t-6 \right)+11\left( t-6 \right)=0 \\
 & \Rightarrow \left( t-6 \right)\left( 2t+11 \right)=0.........equation(ii) \\
\end{align}\]
We know that if \[ab=0\] then either \[a=0\] or \[b=0\]
By using the above result to equation (ii) let us take the first term then we get
\[\Rightarrow t=6\]
By substituting \[\sqrt{3{{x}^{2}}-8x+1}=t\] and squaring on both sides we get
\[\begin{align}
  & \Rightarrow \sqrt{3{{x}^{2}}-8x+1}=6 \\
 & \Rightarrow 3{{x}^{2}}-8x+1=36 \\
 & \Rightarrow 3{{x}^{2}}-8x-35=0 \\
\end{align}\]
Now, by using the factorisation method we get
\[\begin{align}
  & \Rightarrow 3{{x}^{2}}-15x+7x-35=0 \\
 & \Rightarrow 3x\left( x-5 \right)+7\left( x-5 \right)=0 \\
 & \Rightarrow \left( x-5 \right)\left( 3x+7 \right)=0 \\
\end{align}\]
We know that if \[ab=0\] then either \[a=0\] or \[b=0\]
By using the above result we get
\[\Rightarrow x=5,\dfrac{-7}{3}\]
Similarly by taking the second term in equation (ii) we get
\[\Rightarrow t=-\dfrac{11}{2}\]
By substituting \[\sqrt{3{{x}^{2}}-8x+1}=t\] and squaring on both sides we get
\[\begin{align}
  & \Rightarrow \sqrt{3{{x}^{2}}-8x+1}=-\dfrac{11}{2} \\
 & \Rightarrow 3{{x}^{2}}-8x+1=\dfrac{121}{4} \\
 & \Rightarrow 12{{x}^{2}}-32x-117=0 \\
\end{align}\]
We know that the formula of roots of quadratic equation \[a{{x}^{2}}+bx+c=0\] is given as
\[\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Now by using the above formula we get
\[\begin{align}
  & \Rightarrow x=\dfrac{32\pm \sqrt{{{32}^{2}}-4\times 12\left( -117 \right)}}{2\times 12} \\
 & \Rightarrow x=\dfrac{32\pm \sqrt{1024+5616}}{24} \\
 & \Rightarrow x=\dfrac{32\pm \sqrt{6640}}{24} \\
\end{align}\]
Now, by replacing \[\sqrt{6640}\] by \[4\sqrt{415}\] we get
\[\begin{align}
  & \Rightarrow x=\dfrac{32\pm 4\sqrt{415}}{24} \\
 & \Rightarrow x=\dfrac{8\pm \sqrt{415}}{6} \\
\end{align}\]
Therefore the values of \['x'\] satisfying the given equation are
\[\therefore x=5,\dfrac{-7}{3},\dfrac{8+\sqrt{415}}{6},\dfrac{8-\sqrt{415}}{6}\]

Note: Students may make mistakes in solving the problem by using the long and difficult method.
Here, we have the equation
\[\Rightarrow 2\left( 3{{x}^{2}}-8x-32 \right)=\sqrt{3{{x}^{2}}-8x+1}\]
Now, we can solve this problem with squares on both sides. But it results in the equation of \['x'\] having the degree of 4 which will be a difficult and long way of solving.
But we can solve in an easy method by converting the equation in such a way that we get the common term of \[\sqrt{3{{x}^{2}}-8x+1}\] as follows
\[\Rightarrow 2\left( {{\left( \sqrt{3{{x}^{2}}-8x+1} \right)}^{2}}-33 \right)=\sqrt{3{{x}^{2}}-8x+1}\]
So, here we can replace \[\sqrt{3{{x}^{2}}-8x+1}=t\] to solve the problem easily.