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

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

& \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{aligned}$$ Now, by rearranging the terms in the above equation we get $$\begin{aligned} & \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{aligned}$$ Now, let us get the common term that is $$\sqrt{3{{x}^{2}}-8x+1}$$ on both sides we get $$\begin{aligned} & \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{aligned}$$ Let us assume that $$\Rightarrow \sqrt{3{{x}^{2}}-8x+1}=t$$ By substituting the above equation in equation (i) we get $$\begin{aligned} & \Rightarrow 2\left( {{t}^{2}}-33 \right)=t \\\ & \Rightarrow 2{{t}^{2}}-t-66=0 \\\ \end{aligned}$$ By using the factorisation method to the above equation we get $$\begin{aligned} & \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{aligned}$$ We know that if $$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{aligned} & \Rightarrow \sqrt{3{{x}^{2}}-8x+1}=6 \\\ & \Rightarrow 3{{x}^{2}}-8x+1=36 \\\ & \Rightarrow 3{{x}^{2}}-8x-35=0 \\\ \end{aligned}$$ Now, by using the factorisation method we get $$\begin{aligned} & \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{aligned}$$ We know that if $$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{aligned} & \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{aligned}$$ We know that the formula of roots of quadratic equation $$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{aligned} & \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{aligned}$$ Now, by replacing $$\sqrt{6640}$$ by $$4\sqrt{415}$$ we get $$\begin{aligned} & \Rightarrow x=\dfrac{32\pm 4\sqrt{415}}{24} \\\ & \Rightarrow x=\dfrac{8\pm \sqrt{415}}{6} \\\ \end{aligned}$$ Therefore the values of $$'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.