Question

Find the value of x by solving the below equation:
$\left( \dfrac{1200}{x}+2 \right)(x-10)-1200=60$

Answer 1

Hint: To solve the above equation, we will first expand the terms given in the brackets by multiplying them [that is evaluating $\left( \dfrac{1200}{x}+2 \right)(x-10)$]. We will then solve the resulting equation (which would be in terms of x) to get the value of x. We will then use the formula for finding the roots of the quadratic equation ($a{{x}^{2}}+bx+c=0$), given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to get the answer.

Complete step-by-step solution -
Now, to evaluate the above given equation, we start by multiplying the terms in the brackets of the LHS term. Thus, we have,
$\begin{align}
  & \Rightarrow 1200-\dfrac{12000}{x}+2x-20-1200=60 \\
 & \Rightarrow 2x-\dfrac{12000}{x}=80 \\
\end{align}$
Now, to solve this further, we multiply LHS and RHS by x. Thus, we get,
$\begin{align}
  & \Rightarrow 2{{x}^{2}}-12000=80x \\
 & \Rightarrow {{x}^{2}}-6000=40x \\
\end{align}$
Now, subtract 40x from LHS and RHS, we get,
$\Rightarrow {{x}^{2}}-40x-6000=0$
Now, we use the formula for finding the roots of the quadratic equation ($a{{x}^{2}}+bx+c=0$), given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Thus, we will use this to find the roots. Since, in this case, a = 1, b = -40 and c = -6000. Thus, we have,
x = $\dfrac{40\pm \sqrt{{{(-40)}^{2}}-4(1)(-6000)}}{2(1)}$
x = $\dfrac{40\pm \sqrt{25600}}{2}$ = $\dfrac{40\pm 160}{2}$
x = 100, -60.
Hence, the values of x are 100 and -60.

Note: One should note that while solving equations, we should be careful of the values which cannot be possible as a solution. For example, in this problem, while solving the equation, one should note that x cannot be 0 (even if we would have got x = 0 as one of the solutions). This is because if we put x = 0 in the expression $\left( \dfrac{1200}{x}+2 \right)(x-10)-1200$, denominator of the term $\dfrac{1200}{x}$ would be 0. This is not possible and thus x = 0 can never be part of the solution (even if one may get that as a solution after solving).