Question

(a+b)²x²-4abx-(a-b)² solve through factorisation method​. find roots a & b are real numbers.
​

Answer 1

By factorisation method :-

(a + b)²x² - 4abx - (a - b)² = 0

→ (a + b)x² - [(a + b)² - (a - b)²]x - (a - b)² = 0

→ (a + b)x² - (a + b)²x + (a - b)²x - (a - b)² = 0

→ [(a + b)²x (x - 1)] + [(a - b)² (x - 1)] = 0

Take (x - 1) as common

→ (x - 1) [(a + b)²x² + (a - b)²] = 0

→ $\boxed{\sf{x\:=\: 1, \: - \bigg(\frac{ a\: - \: b }{a \: + \: b} \bigg)^{2} }}$

By Quadratic formula :-

(a + b)²x² - 4abx - (a - b)² = 0

We know that

x = $\sf{\frac{ - b \: \pm \: \sqrt{ {(b)}^{2} \: - \: 4ac} }{2a}}$

Here,

a = (a + b)², b = -4ab, c = (a - b)²

Substitute them in above formula

→ x = $\sf\frac{ - (-4ab) \: \pm \: \sqrt{ {(-4ab)}^{2} \: - \: 4(a\:+\:b)^{2}(a\:-\:b)^{2}} }{2(a\:+\:b)^{2}}$

→ $\sf\frac{4ab \: \pm \: \sqrt{16 {a}^{2} {b}^{2} \: + \:4 {a}^{4} \: + \: 4b^{4} \: - \: 8 {a}^{2} {b}^{2} } }{2(a\:+\:b)^{2}}$

→ $\sf\frac{4ab \: \pm \: \sqrt{8 {a}^{2} {b}^{2} \: + \:4 {a}^{4} \: + \: 4b^{4} } }{2(a\:+\:b)^{2}}$

→ $\sf\frac{2(2ab \: \pm \: \sqrt{2{a}^{2} {b}^{2} \: + \: {a}^{4} \: + \: b^{4} } ) }{2(a\:+\:b)^{2}}$

→ $\sf\frac{2ab \: \pm \: \sqrt{2{a}^{2} {b}^{2} \: + \: {a}^{4} \: + \: b^{4} } }{(a\:+\:b)^{2}}$

Used identity: (a² + b²)² = a⁴ + b⁴ + 2a²b²

→ x = $\sf\frac{2ab \: \pm \: \sqrt{( {a^{2} \: + \: {b}^{2})}^{2} } }{(a\:+\:b)^{2}}$

→ x = $\sf\frac{2ab \: \pm \: ( {a^{2} \: + \: {b}^{2})} }{(a\:+\:b)^{2}}$

→ x = $\sf\frac{2ab \: +\: ( {a^{2} \: + \: {b}^{2})} }{(a\:+\:b)^{2}}$

Used identity: (a + b)² = a² + b² + 2ab

→ x = $\sf\frac{ {a}^{2} \: + \: {b}^{2} \: + \: 2ab}{{a}^{2} \: + \: {b}^{2} \: + \: 2ab }$

→ $\boxed{\sf{x\: = \:1}}$

Similarly;

→ x = $\sf\frac{2ab \: -\: ( {a^{2} \: + \: {b}^{2})} }{(a\:+\:b)^{2}}$

→ x = $\sf\frac{ {-\:a}^{2} \: - \: {b}^{2} \: + \: 2ab}{{a}^{2} \: + \: {b}^{2} \: + \: 2ab }$

→ x = $\sf\frac{ - {( a\: - \: b)}^{2} }{( a \: + \: b)^{2} }$

→ $\boxed{\sf{x\:= \: - \bigg(\frac{ a\: - \: b }{a \: + \: b} \bigg)^{2} }}$

Answer 2

||✪✪ QUESTION ✪✪||

(a+b)²x²-4abx-(a-b)² solve through factorisation method. find roots a & b are real numbers.

|| ✰✰ ANSWER ✰✰ ||

Put The Equation to Zero First,

→ (a+b)²x²-4abx-(a-b)² = 0

Dividing both sides by (a+b)² we get,

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

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

Now, adding 4a²b²/(a + b)⁴ both sides we get,

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

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

Now, Comparing LHS, with a² + b² - 2ab = (a - b)² we get ,

→ [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

Square - Root Both Sides now, we get,

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

→ x = [ 2ab ± (a² + b²) ] /(a + b)²

_____________

Taking (+ve) Sign now,

→ x = [2ab + (a² + b²)] / (a+b)²

As , x² + y² + 2xy = (x+y)²

→ x = (a+b)² / (a+b)²

→ x = 1.

______________

Taking (-ve) sign now,

→ x = [ 2ab - (a² + b²) ] /(a+b)²

Taking (-1) common From RHS Numerator,

→ x = (-1)[a² + b² - 2ab] /(a+b)²

Using , x² + y² - 2xy = (x-y)² now,

→ x = [ (-1) (a-b)² / (a+b)² ]

_______________

So, Roots of The Required Equation are 1 and [ (-1) (a-b)² / (a+b)² ] .

And, Factors of The Equation are (x - 1) and [ x + {(a-b)² / (a+b)²} ] .

꧁________________________꧂