Question

Solve by factorisation (a+b)^2x^2-4abx-(a-b)^2=0

Answer 1

(a + b)²x² - 4abx - (a - b)² = 0
⇒ x² - 4abx/(a + b)² - (a - b)²/(a + b)² = 0
⇒ x² - 4ab/(a + b)²x - [(a - b)/(a + b)]² = 0
⇒ x² - 4ab/(a + b)²x + 4a²b²/(a + b)⁴ = [(a - b)/(a + b)]² + 4a²b²/(a + b)⁴
⇒ [x - 2ab/(a + b)²]² = {(a - b)²(a + b)² + 4a²b²}/(a + b)⁴
⇒ [x - 2ab/(a + b)²]² = {(a² + b²)/(a + b)²]²
Taking square root both sides,
⇒[x - 2ab/(a + b)² ] = ± (a² + b²)/(a + b)²
⇒x = {2ab ± (a² + b²)}/(a + b)²
⇒x = (a + b)²/(a + b)² , -[(a -b)/(a + b)]²
⇒x = 1 , -(a - b)²/(a + b)²

Hence, factors are (x - 1) and [x + (a - b)²/(a + b)² ]

Answer 2

Answer:

(a + b)²x² - 4abx - (a - b)² = 0

⇒ x² - 4abx/(a + b)² - (a - b)²/(a + b)² = 0

⇒ x² - 4ab/(a + b)²x - [(a - b)/(a + b)]² = 0

⇒ x² - 4ab/(a + b)²x + 4a²b²/(a + b)⁴ = [(a - b)/(a + b)]² + 4a²b²/(a + b)⁴

⇒ [x - 2ab/(a + b)²]² = {(a - b)²(a + b)² + 4a²b²}/(a + b)⁴

⇒ [x - 2ab/(a + b)²]² = {(a² + b²)/(a + b)²]²

Taking square root both sides,

⇒[x - 2ab/(a + b)² ] = ± (a² + b²)/(a + b)²

⇒x = {2ab ± (a² + b²)}/(a + b)²

⇒x = (a + b)²/(a + b)² , -[(a -b)/(a + b)]²

⇒x = 1 , -(a - b)²/(a + b)²