Question

HEY❤
BRAINLIEST ANSWER

solve the following by factorization...
(a+b)² x² - 4abx - (a-b)² = 0

sol... x = 1 , x = -(a-b)/(a+b)...

acc. to RD Sharma (2013)​

Answer 1

(a + b)²x² - 4abx - (a - b)² = 0

____________ [ GIVEN ]

• We have to find the value of x.

_____________________________

→ (a + b)²x² - 4abx - (a - b)² = 0

We know that..

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

Here ..

a = (a + b)², b = -4ab, c = (a - b)²

Put them in above formula

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

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

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

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

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

Now..

(a² + b²)² = a⁴ + b⁴ + 2a²b²

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

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

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

Now.. (a + b)² = a² + b² + 2ab

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

→ x = 1

Similarly;

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

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

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

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

_____________________________

x = 1, $\frac{ - ( a\: - \: b) }{( a \: + \: b)}$

_________ [ ANSWER ]

_____________________________

Answer 2

SOLUTION

$(a + b) {}^{2} {x}^{2} - 4abx - (a - b) {}^{2} = 0 \\ = > (a + b) {}^{2} {x}^{2} - (a + b) {}^{2} x + (a - b) {}^{2} x - (a - b) {}^{2} = 0 \: \: </u><u>[</u><u>A</u><u>s</u><u> - (a + b) {}^{2} + (a - b) {}^{2} = > - {a}^{2} - {b}^{2} - 2ab + {a}^{2} + {b}^{2} - 2ab = - 4ab</u><u>]</u><u> \\ \\ = > (a + b) {}^{2} \times </u><u>[</u><u>x</u><u> - 1</u><u>]</u><u>+ (a - b) {}^{2} </u><u>[</u><u>x</u><u> - 1</u><u>]</u><u>= 0 \\ \\ = > </u><u>(</u><u>x</u><u> - 1</u><u>)</u><u>[</u><u>(a + b) {}^{2} x + (a - b) {}^{2} </u><u>]</u><u> = 0 \\ \\ = > x - 1 = 0 \: \: or \: \: (a + b) { }^{2} x + (a - b) {}^{2} = 0 \\ \\ = > x = 1 \: \: or \: \: x = - \frac{(a - b) {}^{2} }{(a + b) {}^{2} } \\ \\ = > x = 1 \: \: \: or \: \: \: x = - </u><u>[</u><u>\frac{a - b}{a + b} </u><u>]</u><u> {}^{2}$

Therefore, x has two values:

$= > x = 1 \: \: \: \: or \: \: - (\frac{a - b}{a + b} ) {}^{2}$

Hope it helps ☺️