Question

How do you solve the quadratic \[3+\dfrac{5}{2x}=\dfrac{1}{{{x}^{2}}}\] using any method?

Answer 1

Hint: A quadratic is a polynomial expression like \[a{{x}^{2}}+bx+c\]. We can solve for x by rewriting the equation like \[a{{x}^{2}}+bx+c=0\]. From this equation, we can find x by three methods: finding factors, using quadratic formulas or completing the square. And the quadratic formula is \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].

Complete step by step answer:
As per the given question, we have to solve the given quadratic equation to get the respective x values. We are provided with the quadratic equation \[3+\dfrac{5}{2x}=\dfrac{1}{{{x}^{2}}}\].
In the given equation, we have fractions. In order to get rid of them we have to multiply the whole equation with the LCM of the denominators.
In the equation \[3+\dfrac{5}{2x}=\dfrac{1}{{{x}^{2}}}\] , the denominators of fractions are \[2x\] and \[{{x}^{2}}\]. The LCM of \[2x\] and \[{{x}^{2}}\] is \[2{{x}^{2}}\]. So, we multiply the whole equation with \[2{{x}^{2}}\]. Then, we get \[\Rightarrow 2{{x}^{2}}\times 3+2{{x}^{2}}\times \dfrac{5}{2x}=2{{x}^{2}}\times \dfrac{1}{{{x}^{2}}}\] -----(1)
In equation 1, \[2{{x}^{2}}\times 3\] equals to \[6{{x}^{2}}\], \[2{{x}^{2}}\times \dfrac{5}{2x}\] equals to \[5x\] and \[2{{x}^{2}}\times \dfrac{1}{{{x}^{2}}}\] equals to 2. So, by substituting these values into equation (1), we get
\[\Rightarrow 6{{x}^{2}}+5x-2=0\] ------(2)
Generally, we can solve quadratic equations by finding factors or by completing the square method or by using quadratic formulas.
We can’t find the solution by factoring as there is no pair of numbers whose product is -12 and sum is 5.
So, let’s use the quadratic formula: \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. Here, in equation \[6{{x}^{2}}+5x-2=0\], we have \[a=6\], \[b=5\] and \[c=-2\]. By substituting these values into the formula, we get
\[\begin{align}
  & \Rightarrow x=\dfrac{-5\pm \sqrt{{{(5)}^{2}}-4(6)(-2)}}{2(6)} \\
 & \Rightarrow x=\dfrac{-5\pm \sqrt{25+48}}{12} \\
 & \Rightarrow x=\dfrac{-5\pm \sqrt{73}}{12} \\
\end{align}\]
\[\therefore x=\dfrac{-5\pm \sqrt{73}}{12}\] are the two solutions of the equation \[3+\dfrac{5}{2x}=\dfrac{1}{{{x}^{2}}}\].

Note:
We can solve the quadratic equation \[3+\dfrac{5}{2x}=\dfrac{1}{{{x}^{2}}}\] to get the x values by converting the equation into the form of \[{{(x-h)}^{2}}=k\]. As the simplified form of \[3+\dfrac{5}{2x}=\dfrac{1}{{{x}^{2}}}\] is \[6{{x}^{2}}+5x-2=0\]. This equation is same as \[{{\left( x+\dfrac{5}{12} \right)}^{2}}=\dfrac{73}{144}\] which is easy to solve. We should avoid calculation mistakes.