Question

12. Find the roots of the equation 5x
^2
– 6x – 2 = 0, by method of competing the square.

Standard:- 10

Content Quality Solution Required

No Spamming

Answer 1

Hey!!!

Good Morning

Difficulty Level : Average

Chances of being asked in Board : 80%

_________________

We have

=> 5x² - 6x - 2 = 0

Let's make the co efficient of x² 1

Divide both sides by 5

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

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

Add the square of half of (6/5) on both sides

$= > {x}^{2} - \frac{6}{5} x + { (\frac{3}{5} )}^{2} = \frac{2}{5} + {( \frac{3}{5} )}^{2}$
We know (a - b)² = a² - 2ab + b²

$= > {(x - \frac{3}{5}) }^{2} = \frac{2}{5} + \frac{9}{25}$

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

$= > x + \frac{3}{5} = + - \sqrt{ \frac{19}{25} }$

$= > x + \frac{3}{5} = + - \frac{ \sqrt{19} }{5}$

Thus

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

Hope this helps ✌️