Question

Solve the given expression by the method of completing the square $5{{x}^{2}}-6x-2=0$
(a) $\dfrac{3+\sqrt{19}}{5}and\dfrac{3-\sqrt{19}}{5}$
(b) $\dfrac{-3+\sqrt{19}}{5}and\dfrac{3-\sqrt{19}}{5}$
(c) $\dfrac{3+\sqrt{19}}{3}and\dfrac{3-\sqrt{19}}{5}$
(d) $\dfrac{3+\sqrt{19}}{5}and\dfrac{-3-\sqrt{19}}{3}$

Answer 1

Hint: Transfer constant term to other side of the given equation. Now divide the whole equation by the coefficient of ${{x}^{2}}$ . And observe the algebraic identity of ${{\left( a-b \right)}^{2}}$ , given as
${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
Compare the terms of L.H.S. of calculated expression by ${{a}^{2}}$ and $-2ab$ and get the value of b. Add ${{b}^{2}}$ to both sides of the expression and write the L.H.S in square form (Add ${{\left( \dfrac{3}{5} \right)}^{2}}$ to both sides). Now, take square root to both the sides to get roots.

Complete step-by-step solution -
As we know the completing square method to get roots of any quadratic equation tells us that we can get roots by making the given quadratic in perfect square form using the variable terms only i.e. terms with ${{x}^{2}}$ and $x$.
So, given quadratic in the problem is
$5{{x}^{2}}-6x-2=0............\left( i \right)$
Transfer the term -2 to other side of the equation, we get
$5{{x}^{2}}-6x=2$
Divide the whole equation by 5, we get
$\begin{aligned} & \dfrac{5{{x}^{2}}-6x}{5}=\dfrac{2}{5} \\\ & \Rightarrow \dfrac{5{{x}^{2}}}{5}-\dfrac{6x}{5}=\dfrac{2}{5} \\\ & {{x}^{2}}-\dfrac{6x}{5}=\dfrac{2}{5}..................\left( ii \right) \\\ \end{aligned}$
Now, as we know the algebraic identity of ${{\left( a-b \right)}^{2}}$ is given as
${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}..............\left( iii \right)$
Now, compare the L.H.S. of the equation (ii) and R.H.S of equation (iii) (only first two terms). So, we get that ${{x}^{2}}$ as acting as ${{a}^{2}}$ and $2ab$ is acting as $\dfrac{6x}{5}$ i.e. $2\times x\times \dfrac{3}{5}$ .So, if we want to make L.H.S. of equation (ii) as perfect square, then we need to add ${{\left( \dfrac{3}{5} \right)}^{2}}$ to it according to the comparison of equation (iii).
Hence, adding ${{\left( \dfrac{3}{5} \right)}^{2}}$ to both sides of equation (ii), we get equation (ii) as
$\begin{aligned} & {{x}^{2}}-\dfrac{6x}{5}+{{\left( \dfrac{3}{5} \right)}^{2}}=\dfrac{2}{5}+{{\left( \dfrac{3}{5} \right)}^{2}} \\\ & \Rightarrow {{\left( x \right)}^{2}}-2\times x\times \dfrac{3}{5}+{{\left( \dfrac{3}{5} \right)}^{2}}=\dfrac{2}{5}+\dfrac{9}{25} \\\ \end{aligned}$
On comparing the L.H.S of above equation and R.H.S of equation (iii), we get
$a=x$ and $b=\dfrac{3}{5}$
So, we can replace the L.H.S. of above equation as ${{\left( x-\dfrac{3}{5} \right)}^{2}}$ using L.H.S. of equation (iii). So, we get
$\begin{aligned} & {{\left( x-\dfrac{3}{5} \right)}^{2}}=\dfrac{10+9}{25}=\dfrac{19}{25} \\\ & {{\left( x-\dfrac{3}{5} \right)}^{2}}=\dfrac{19}{25} \\\ \end{aligned}$
Now, taking square root to both sides of the equation, we can rewrite the equation as
$\begin{aligned} & \sqrt{{{\left( x-\dfrac{3}{5} \right)}^{2}}}=\pm \sqrt{\dfrac{19}{25}} \\\ & \Rightarrow x-\dfrac{3}{5}=\pm \sqrt{\dfrac{19}{25}} \\\ \end{aligned}$
Transferring $\dfrac{3}{5}$ to other side of the equation, we can get values of $x$ as
$\begin{aligned} & x=\dfrac{3}{5}\pm \dfrac{\sqrt{19}}{5} \\\ & \Rightarrow x=\dfrac{3\pm \sqrt{19}}{5} \\\ \end{aligned}$
So, both the roots individually are given as
$x=\dfrac{3+\sqrt{19}}{5}\text{ }and\text{ }\dfrac{3-\sqrt{19}}{5}$
Hence option (a) is the correct answer.

Note: Another approach to get the roots of the quadratic would be that we can use quadratic formula, if we are not restricted to use only one method. It is given for a standard quadratic i.e. $a{{x}^{2}}+bx+c=0$ , as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Put a = 5, b = -6, c = -2 to get the roots of the given quadratic.
The above formula is proved by the method given in the problem. So, getting roots by using formation of square is the fundamental approach to get roots. It is given as

& a{{x}^{2}}+bx+c=0 \\\ & \Rightarrow {{x}^{2}}+\dfrac{b}{a}x=\dfrac{-c}{2} \\\ & {{\left( x+\dfrac{b}{2a} \right)}^{2}}=\dfrac{-c}{a}+\dfrac{{{b}^{2}}}{4{{a}^{2}}} \\\ & \Rightarrow {{\left( x+\dfrac{b}{2a} \right)}^{2}}=\dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}} \\\ & x=\dfrac{-b}{2a}\pm \sqrt{\dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}}} \\\ & \Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\\ \end{aligned}$$ So, formation of square by using terms with coefficients ${{x}^{2}}$ and $x$ . Observation is the key for writing them in square form using the identities ${{\left( a+b \right)}^{2}}$ or ${{\left( a-b \right)}^{2}}$ .Follow the above approach for getting roots of any quadratic using method square formation.