Question

solve for x. please please please​

Answer 1

Answer:

compare the given equation with general equation of ax^2 + bx +c=0

here the roots let it be x= {-b+_ (b^2-4ac)^1/2}/2a

Step-by-step explanation:

Answer 2

(a + b)²x² - 4abx - (a - b)² = 0

____________ [ GIVEN ]

• We have to find the value of x.

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

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

_________ [ ANSWER ]

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