Question

Example 8: Find the roots of the equation 5x? - 6x - 2 = 0 by the method of completing
the square

Answer 1

$\large\bf\underline{Given:-}$

p(x) = 5x² - 6x -2 = 0

$\large\bf\underline {To \: find:-}$

• root's of the given equation by completing the square method

$\huge\bf\underline{Solution:-}$

• equation :- 5x² - 6x -2 = 0

Divide both side of equation by 5.

$\rm \longmapsto \: \frac{ {5x}^{2} }{5} - \frac{6x}{5} - \frac{2}{5} = \frac{0}{5} \\ \\ \rm \longmapsto \: {x}^{2} - \frac{6x}{5} - \frac{2}{5} = 0 \\ \\ \rm \longmapsto \: {x}^{2} - \frac{6x}{5} = \frac{2}{5}$

Add (6/10)² = (3/5)²on both sides of the equation.

$\rm \longmapsto \: {x}^{2} - \frac{6x}{5} + \frac{9}{25} = \frac{2}{5} + \frac{9}{25} \\ \\ \rm \longmapsto \:( x + \frac{3}{5} ) {}^{2} = \frac{10 + 9}{25} \\ \\ \rm \longmapsto \: (x + \frac{3}{5} ) {}^{2} = \frac{19}{25} \\ \\ \rm \longmapsto \: x + \frac{3}{5} = \sqrt{ \frac{19}{25} } \\ \\ \rm \longmapsto \: x = \frac{ \pm \sqrt{ 19} }{5} - \frac{3}{5} \\ \\ \rm \longmapsto \: x = \frac{ \sqrt{19} - 3 }{5} \: or \: x = \frac{ - ( \sqrt{19} + 3)}{5}$

So ,the roots of the given equation are as :-

$\bf \: x = \frac{ \sqrt { 19 }- 3}{5} \: or \: x = \frac{ - ( \sqrt{19} + 3) }{5}$

Answer 2

$\sf\blue{Correct \ question}$

$\sf{Find \ the \ roots \ of \ the \ equation}$

$\sf{5x^{2}-6x-2=0 \ by \ the \ method \ of}$

$\sf{completing \ the \ square.}$

___________________________________

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{\frac{-3+\sqrt19}{5} \ and \ \frac{-3-\sqrt19}{5} \ are \ the \ roots \ of \ the \ equation.}$

$\sf\orange{\underline{\underline{Given:}}}$

$\sf{The \ given \ quadratic \ equation \ is}$

$\sf{\implies{5x^{2}-6x-2=0}}$

$\sf\pink{To \ find:}$

$\sf{Roots \ of \ the \ equation.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{The \ given \ quadratic \ equation \ is}$

$\sf{\implies{5x^{2}-6x-2=0}}$

$\sf{Divide \ the \ equation \ through \ by \ 5}$

$\sf{\implies{x^{2}-\frac{6x}{5}-\frac{2}{5}=0}}$

$\sf{\implies{x^{2}-\frac{6x}{5}=\frac{2}{5}}}$

__________________________________

$\sf{(\frac{1}{2}\times \ Coefficient \ of \ x)^{2}}$

$\sf{\implies{(\frac{1}{2}\times\frac{6}{5})^{2}}}$

$\sf{\implies{(\frac{3}{5})^{2}}}$

$\sf{\implies{\frac{9}{25}}}$

________________________________

$\sf{Add \ \frac{9}{25} \ on \ both \ sides \ of \ the \ equation}$

$\sf{\implies{x^{2}+\frac{6x}{5}+\frac{9}{25}=\frac{2}{5}+\frac{9}{25}}}$

$\sf{\implies{x^{2}+\frac{6x}{5}+\frac{9}{25}=\frac{10+9}{25}}}$

$\sf{\implies{(x+\frac{3}{5})^{2}=\frac{19}{25}}}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\sf{\implies{x+\frac{3}{5}=\frac{\sqrt19}{5} \ or \ -\frac{\sqrt19}{5}}}$

$\sf{\implies{x=\frac{-3}{5}+\frac{\sqrt19}{5} \ or \ \frac{-3}{5}-\frac{\sqrt19}{5}}}$

$\sf{\implies{x=\frac{-3+\sqrt19}{5} \ or \ \frac{-3-\sqrt19}{5}}}$

$\sf\purple{\tt{\therefore{\frac{-3+\sqrt19}{5} \ and \ \frac{-3-\sqrt19}{5} \ are \ the \ roots \ of \ the \ equation.}}}$