Question

MATHEMATICS
LE 3
Find the roots of the equation 5x2 - 6x - 2 = 0 by the method of completine
[NCERT​

Answer 1

Q: Find the roots of the equation 5x² - 6x - 2 = 0 by the method of completing square.

Solution :-

Q : 5x² - 6x - 2 = 0

=> 5x² - 6x = 2

Dividing the equation by 5 :

=> x² - 6x/5 = 2/5

Adding (6/5 × 1/2)² both sides :

=> x² - 6x/5 + (3/5)² = 2/5 + (3/5)²

=> (x - 3/5)² = 2/5 + 9/25

=> (x - 3/5) (x + 3/5) = (10 + 9)/25

=> x = 19/25 + 3/5 or x = 19/25 - 3/5

=> x = (19 + 15)/25 or x = (19 - 15)/25

=> x = 34/25 or x = 4/25

Answer : Value of x = 34/25 or 4/25

Answer 2

5x² - 6x - 2 = 0

___________ [ GIVEN ]

• We have to find the roots if the equation by completing the square method.

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→ 5x² - 6x - 2 = 0

Divide throughout with 5

→ $\dfrac{5{x}^{2}}{5}$ - $\dfrac{6x}{5}\:-\:\dfrac{2}{5}$ = 0

→ x² - $\dfrac{6x}{5}\:-\:\dfrac{2}{5}$ = 0

→ x² - $\dfrac{6x}{5}\:=\:\dfrac{2}{5}$

Now..

→ x² - 2 × $\dfrac{3x}{5}$ + $\bigg(\dfrac{3}{5}\bigg)^{2}$ = $\dfrac{2}{5}$ + $\bigg(\dfrac{3}{5}\bigg)^{2}$

→ $\bigg(x\:-\:\dfrac{3}{5}\bigg)^{2}$ = $\dfrac{19}{25}$

→ x - $\dfrac{3}{5}$ = ± $\sqrt{\dfrac{19}{25}}$

→ x - $\dfrac{3}{5}$ = ± $\dfrac{\sqrt{19}}{5}$

→ x = ± $\dfrac{\sqrt{19}}{5}$ + $\dfrac{3}{5}$

→ x = $\dfrac{3\:\pm\:\sqrt{19}}{5}$

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

__________ [ ANSWER ]

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