Question

Hi friends

Solve by completing the square

$\sqrt {2} x^2-3x-2\sqrt{2}=0$

Answer in book = -1/root2 and 2root2

NO SPAM PLEASE

Answer 1

Answer:

-1/√2, 2√2

Step-by-step explanation:

Given Equation is √2x² - 3x - 2√2 = 0.

⇒ √2x² - 3x = 2√2

(1) Coefficient of x² should be 1:

⇒ x² - (3/√2)x = 2.

(2) Add the square of half the coefficient of 'x' on both sides:

x² - (3/√2) + (3/2√2)² = 2 + (3/2√2)²

x² - (3/√2) + (9/8) = 2 + (3/2√2)²

(3) Three terms on Left side will be a² + b² + 2ab:

⇒ (x - 3/2√2)² = 2 + (9/8)

⇒ (x - 3/2√2)² = 25/8

⇒ (x - 3/2√2) = ±√25/8

(4) Solve for x by simplification:

(i)

When x = +√25/8:

⇒ (x - 3/2√2) = √25/8

⇒ x - (3√2/4) = (5√2)/4

⇒ x = (5√2/4) + (3√2/4)

⇒ x = 8√2/4

⇒ x = 2√2

(ii)

When x = -√25/8.:

⇒ (x - 3/2√2) = -(√25/8)

⇒ x - (3√2/4) = -(5√2)/4

⇒ x = -(5√2/4) + (3√2/4)

⇒ x = -√2/2

⇒ x = -1/√2.

Therefore, x = -1/√2, 2√2 are the roots of the equation.

Hope it helps!

Answer 2

hey mate

here is your answer


thank you for posting your question!