Question

Solve the following equation \[x-\dfrac{18}{x}=6\], give your answer correct to two significant figures.

Answer 1

Hint: In this problem we have to solve the given quadratic equation and find the value of x. Here we can first form a quadratic equation by cross multiplication. We know that to solve a quadratic equation, we can use two methods, the one is the quadratic formula method and the other is the factorisation method. To solve by using quadratic formula method, we have to use the quadratic formula\[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] and substitute the value of a, b, c from the given equation to find the value of x.

Complete step by step answer:
We know that the given equation is,
\[x-\dfrac{18}{x}=6\]
We can now cross multiply and simplify the given equation, we get
\[\Rightarrow {{x}^{2}}-6x-18\]…….. (1)
We also know that a quadratic equation in standard form is,
\[a{{x}^{2}}+bx+c=0\] ……. (2)
We can now compare the two equations (1) and (2), we get
a = 1, b = -6, c = -18.
We know that the quadratic formula for the standard form \[a{{x}^{2}}+bx+c=0\] is
\[x=\dfrac{-b\pm \sqrt{{{\left( b \right)}^{2}}-4\times a\times \left( c \right)}}{2a}\]
Now we can substitute the value of a, b, c in the above formula, we get
\[\Rightarrow x=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\times 1\times \left( -18 \right)}}{2\times 1}\]
Now we can simplify the above step, we get
\[\begin{align}
  & \Rightarrow x=\dfrac{\left( 6 \right)\pm \sqrt{36+72}}{2\times 1} \\
 & \Rightarrow x=\dfrac{6\pm \sqrt{108}}{2} \\
\end{align}\]
Now we can separate the terms to simplify it,
\[\begin{align}
  & \Rightarrow x=\dfrac{6+10.39}{2}=\dfrac{16.39}{2}=8.19 \\
 & \Rightarrow x=\dfrac{6-10.39}{2}=\dfrac{-4.29}{2}=-2.19 \\
\end{align}\]
Therefore, the value of \[x=8.19\] and \[x=-2.19\].

Note: We can also use a simple factorisation method to solve this problem but we have to know that the answer is in root form, so we can solve only by using a quadratic formula to get the final answer correctly. Students may make mistakes in the quadratic formula part, which should be concentrated. we should also concentrate while substituting the values in the quadratic formula.