Question

Find the roots of the equation $5{{x}^{2}}-6x-2=0$, by the method of completing the square.

Answer 1

Hint: In this given question, firstly we should make the term with ${{x}^{2}}$ a square term, here we may multiply or divide the whole equation by 5 to do this. Then we must represent the term with $x$ as twice the product of those two terms of which we want to square by multiplying and dividing it with 2. Then we should add and subtract the square of the constant term in the above step. Thereafter, we must simplify the equation in order to get our value for $x$ as the roots of this equation.

Complete step-by-step answer:

In this given question, we are asked to find the roots of the given equation, that is:

$5{{x}^{2}}-6x-2=0$………… (1.1), by the method of completing the square.

Basically, here we are going to represent this equation as the square of a linear equation with some extra addition in it.

In order to do so and get our answer, we should follow the following steps:

Let’s divide the whole equation (1.1) by 5, to get

$\begin{align}

  & \dfrac{5{{x}^{2}}-6x-2}{5}=\dfrac{0}{5} \\

 & \Rightarrow {{x}^{2}}-\dfrac{6}{5}x-\dfrac{2}{5}=0............(1.2) \\

\end{align}$

Now,

$\begin{align}

  & {{x}^{2}}-\dfrac{6}{5}x-\dfrac{2}{5}=0 \\

 & \Rightarrow {{x}^{2}}-2\times x\times \dfrac{3}{5}-\dfrac{2}{5}=0...........(1.3) \\

\end{align}$

Then let’s add and subtract ${{\left( \dfrac{3}{5} \right)}^{2}}$ from the Left Hand Side (LHS) of equation 1.3, to get

\[\begin{align}

  & {{x}^{2}}-2\times x\times \dfrac{3}{5}-\dfrac{2}{5}+{{\left( \dfrac{3}{5} \right)}^{2}}-{{\left( \dfrac{3}{5} \right)}^{2}}=0 \\

 & \Rightarrow {{x}^{2}}-2\times x\times \dfrac{3}{5}+{{\left( \dfrac{3}{5} \right)}^{2}}-\dfrac{2}{5}-{{\left( \dfrac{3}{5} \right)}^{2}}=0 \\

 & \Rightarrow {{\left( x-\dfrac{3}{5} \right)}^{2}}={{\left( \dfrac{3}{5} \right)}^{2}}+\dfrac{2}{5} \\

 & \Rightarrow {{\left( x-\dfrac{3}{5} \right)}^{2}}=\dfrac{9}{25}+\dfrac{2}{5} \\

 & \Rightarrow {{\left( x-\dfrac{3}{5} \right)}^{2}}=\dfrac{19}{25} \\

 & \Rightarrow x-\dfrac{3}{5}=\pm \sqrt{\dfrac{19}{25}} \\

 & \Rightarrow x=\dfrac{3}{5}\pm \dfrac{\sqrt{19}}{5} \\

 & \Rightarrow x=\dfrac{3\pm \sqrt{19}}{5}.............(1.4) \\

\end{align}\]

So, we get the roots of the given equation as

$x=\dfrac{3+\sqrt{19}}{5}$ and $x=\dfrac{3-\sqrt{19}}{5}$.

Therefore, our solution is done and we have obtained the roots of the equation $5{{x}^{2}}-6x-2=0$ as our required answer.

Note: In this question, we may have also obtained the same answer if we would have multiplied the whole equation (1.1) by 5 instead of dividing it by 5. Although the process and calculations may have been somewhat different.