Question

Solve the following equation $\dfrac{x}{2}-\dfrac{4}{x}+\dfrac{x}{5}+\dfrac{3x}{10}=\dfrac{1}{3}$ \[\]

Answer 1

Hint: We add the expressions in fractional form at the left hand side of the equation by first finding the least common multiple of the denominators $2,x,5,10$ and then replacing the fractional terms $\dfrac{x}{2},\dfrac{4}{x},\dfrac{x}{5},\dfrac{3x}{10}$ with their equivalent fractional term. We simplify until we get a quadratic equation which we solve by taking the square root of both sides.

Complete step-by-step answer:
We know that when we are asked to solve and given an equation involving unknown variables then we have to find the value of unknown variables. Here in the question the unknown variable is $x$ and the expression is
\[\dfrac{x}{2}-\dfrac{4}{x}+\dfrac{x}{5}+\dfrac{3x}{10}=\dfrac{1}{3}.....\left( 1 \right)\]
So we find the value of $x.$ Let us start by adding the expressions in fraction at the left hand side of the above equation. So we first have to find the least common multiple (LCM) of the denominators$2,5,x,10$. If we just find the LCM of 2,5,10 and then multiply it by $x$we are going to find the LCM of$2,5,x,10$. We proceed as
\[\begin{matrix}
   2\left| \!{\underline {\,
  2,5,10 \,}} \right. \\
   5\left| \!{\underline {\,
  1,5,5 \,}} \right. \\
   1,1,1 \\
\end{matrix}\]
So the $\text{LCM}\left( 2,5,10 \right)=10$ and hence $\text{LCM}\left( 2,5,x,10 \right)=10\times x=10x$. So let us convert each fractional term to the equivalent fraction with the same denominator in the left hand side of the equation (1) . We get
\[ \begin{align}
  & \dfrac{x}{2}=\dfrac{x\times 5x}{2\times 5x}=\dfrac{5{{x}^{2}}}{10x} \\
 & \dfrac{4}{x}=\dfrac{4\times 10}{x\times 10}=\dfrac{40}{10x} \\
 & \dfrac{x}{5}=\dfrac{x\times 2x}{5\times 2x}=\dfrac{2{{x}^{2}}}{10x} \\
 & \dfrac{3x}{10}=\dfrac{3x\times x}{10\times x}=\dfrac{3{{x}^{2}}}{10x} \\
\end{align}\]
We replace the fractional expression with their equivalent fractional expression in equation (1) to have
\[ \begin{align}
  & \dfrac{5{{x}^{2}}}{10x}-\dfrac{40}{10x}+\dfrac{2{{x}^{2}}}{10x}+\dfrac{3{{x}^{2}}}{10x}=\dfrac{1}{3} \\
 & \Rightarrow \dfrac{5{{x}^{2}}+2{{x}^{2}}+3{{x}^{2}}-40}{10}=\dfrac{1}{3} \\
 & \Rightarrow \dfrac{10{{x}^{2}}-40}{10}=\dfrac{1}{3} \\
\end{align}\]
We divide the numerator and denominator of the left hand side by 10 and then cross multiply to have
\[\begin{align}
  & \Rightarrow \dfrac{{{x}^{2}}-4}{1}=\dfrac{1}{3} \\
 & \Rightarrow 3\left( {{x}^{2}}-4 \right)=1 \\
 & \Rightarrow 3{{x}^{2}}-12=1 \\
 & \Rightarrow 3{{x}^{2}}=13 \\
 & \Rightarrow {{x}^{2}}=\dfrac{13}{3} \\
\end{align}\]
We take square root both side of the above quadratic equation and get two values for $x$ as
\[\Rightarrow x=\sqrt{\dfrac{13}{3}},-\sqrt{\dfrac{13}{3}}\]

Note: We know that the quadratic equation in the form $a{{x}^{2}}+bx+c=0$ will have real and distinct roots only when the discriminant $D={{b}^{2}}-4ac > 0$. We can check it in the obtained quadratic equation$3{{x}^{2}}-13=0$. We can alternatively solve the quadratic equation using the splitting the middle term or the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.