Question

How‌ ‌do‌ ‌you‌ ‌solve‌ ‌$\dfrac{1}{x}+\dfrac{1}{x+3}=2$?‌

Answer 1

Hint: First of all simplify L.H.S by taking the L.C.M. Cross multiply the terms to get a quadratic equation in x. Now, take all the terms to the L.H.S and make the coefficient of ${{x}^{2}}$ positive. Use the quadratic formula to solve a quadratic equation given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where ‘a’ is the coefficient of ${{x}^{2}}$, b is the coefficient of x and c is the constant term.

Complete step-by-step solution:
Here we have been provided with the equation $\dfrac{1}{x}+\dfrac{1}{x+3}=2$ and we are asked to solve this equation. That means we have to find the value of x. First let us simplify the L.H.S, so taking the L.C.M we get,
$\begin{align}
  & \Rightarrow \dfrac{x+3+x}{x\left( x+3 \right)}=2 \\
 & \Rightarrow \dfrac{2x+3}{x\left( x+3 \right)}=2 \\
\end{align}$
Cross multiplying the terms we get,
$\begin{align}
  & \Rightarrow 2x+3=2x\left( x+3 \right) \\
 & \Rightarrow 2x+3=2{{x}^{2}}+6x \\
\end{align}$
Taking all the terms to the L.H.S we get,
$\begin{align}
  & \Rightarrow -2{{x}^{2}}+2x-6x+3=0 \\
 & \Rightarrow -2{{x}^{2}}-4x+3=0 \\
\end{align}$
Multiplying both the sides with -1 to make the coefficient of ${{x}^{2}}$ positive we get,
$\Rightarrow 2{{x}^{2}}+4x-3=0$
Considering the above quadratic equation of the form $a{{x}^{2}}+bx+c=0$ and comparing the coefficients we have a = 2, b = 4 and c = -3. Therefore applying the quadratic formula given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and substituting the assumed values we get,
$\begin{align}
  & \Rightarrow x=\dfrac{-4\pm \sqrt{{{4}^{2}}-4\left( 2 \right)\left( -3 \right)}}{2\left( 2 \right)} \\
 & \Rightarrow x=\dfrac{-4\pm \sqrt{40}}{4} \\
 & \Rightarrow x=\dfrac{-4\pm 2\sqrt{10}}{4} \\
\end{align}$
(i) Considering the positive sign we get,
$\begin{align}
  & \Rightarrow x=\dfrac{-4+2\sqrt{10}}{4} \\
 & \Rightarrow x=\dfrac{-2+\sqrt{10}}{2} \\
\end{align}$
(ii) Considering the negative sign we get,
$\begin{align}
  & \Rightarrow x=\dfrac{-4-2\sqrt{10}}{4} \\
 & \Rightarrow x=\dfrac{-2-\sqrt{10}}{2} \\
 & \Rightarrow x=\dfrac{-\left( 2+\sqrt{10} \right)}{2} \\
\end{align}$
Hence, the above two values of x are the solutions of the given equation.

Note: Note that here it will be difficult to use the middle term split method because the values of x in the solutions have radical terms which will be difficult to think of while splitting the middle term. You must remember the quadratic formula to solve the above question. Note that it is not necessary to make the coefficient of ${{x}^{2}}$ positive, you can directly use the quadratic formula.