Question

find the roots of equation 5x²-6x-2=0 by method of completing square​

Answer 1

Answer:

Step-by-step explanation:

5x² - 6x - 2 = 0

dividing the eq by 5

Than we get

X² - 6x/5 - 2/5 = 0

(x) ² - 2× x × 3/5 + (3/5)² - 2/5 - 9/25 = = 0

(x - 3/5)² = 2/5 + 9/25

(x - 3/5)² = 10 + 9 / 25

(x - 3/5)² = 19 / 25

X - 3/5 = /19 ÷25

X-3/5 = ±/19 ÷ 5

x-3/5 = /19 ÷5. , x-3/5 = -/19 ÷5

X = /19 ÷5 + 3÷5. , X = -/19÷5 + 3÷5

X = /19 + 3 ÷ 5. , X = -/19 + 3 ÷ 5

Hope it helps uh

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Answer 2

$<b>Given</b>$ : 5x²-6x-2 = 0

$<b>Step-1 </b>$:

Dividing both sides by 5

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

$<b>Step-2</b>$ :

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

$<b>Step-3</b>$ :

$\sf\:x {}^{2} - \frac{6}{5} x + ( \frac{3}{5} ) {}^{2} = \sf\frac{2}{5} + ( \frac{3}{5} ) {}^{2}$

$<b>Step-4</b>$ :

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

$<b>Step-5</b>$ :

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

$\sf\:x = \frac{3 + \sqrt{19} }{5} \: \: or \: \: \: \: x = \frac{3 - \sqrt{19} }{5}$