Question

✬ HeY There ! ✬

➥ Solve the following equations.

1.} 4x² – 4ax + (a² – b²) = 0
2.} 9x² – 6b²x – (a⁴ – b⁴) = 0​

Answer 1

Answer :-

$\;\underbrace{\underline{\sf{Question's\;\;Analysis\;\;:-}}}$

Here the concept of Quadratic Equations and Algebraic Identities has been. First we will split each and every equation in simplest term and then we will apply identities. After applying that, we will find the value of x. Since, this is a quadratic equation, we will get two values of x.

Let's do it !!

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★ Solution :-

Given quadratic equations :-

1.] 4x² - 4ax + (a² - b²) = 0

2.] 9x² - 6b²x - (a⁴ - b⁴) = 0

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1.) 4x² - 4ax + (a² - b²) = 0

Using identity, a² - b² = (a + b)(a - b), we get

$\\\;\sf{:\rightarrow\;\;\;4x^{2}\;-\;4ax\;+\;[(a\;+\;b)(a\;-\;b)]\;=\;\bf{0}}$

$\\\;\sf{:\rightarrow\;\;\;4x^{2}\;-\;2(a\;+\;b\;+\;a\;-\;b)x\;+\;[(a\;+\;b)(a\;-\;b)]\;=\;\bf{0}}$

$\\\;\sf{:\rightarrow\;\;\;4x^{2}\;-\;2(a\;+\;b\;+\;a\;-\;b)x\;+\;[(a\;+\;b)(a\;-\;b)]\;=\;\bf{0}}$

On simplifying, this we get

$\\\;\sf{:\rightarrow\;\;\;4x^{2}\;-\;2(a\;+\;b)x\;-\;2(a\;-\;b)x\;+\;[(a\;+\;b)(a\;-\;b)]\;=\;\bf{0}}$

$\\\;\sf{:\rightarrow\;\;\;4x^{2}\;-\;2(a\;+\;b)x\;-\;2(a\;-\;b)x\;+\;[(a\;+\;b)(a\;-\;b)]\;=\;\bf{0}}$

Taking 2x as common term, we get,

$\\\;\sf{:\rightarrow\;\;\;2x[2x\;-\;(a\;+\;b)]\;-\;(a\;-\;b)[2x\;-\;[(a\;+\;b)]\;=\;\bf{0}}$

Now taking the common terms as together, we get,

$\\\;\sf{:\rightarrow\;\;\;[2x\;-\;(a\;-\;b)]\:[2x\;-\;[(a\;+\;b)]\;=\;\bf{0}}$

Here either,

[2x - (a - b)] = 0 , or

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

Then,

✒ 2x - (a - b) = 0 or 2x - (a + b) = 0

✒ 2x - a + b = 0 or 2x - a - b = 0

✒ 2x = a - b or 2x = a + b

$\\\;\quad\sf{:\mapsto\;\;x\;=\;\bf{\dfrac{(a\;-\;b)}{2}}\;\;\;\sf{or\;\;\;x\;=}\;\bf{\dfrac{(a\;+\;b)}{2}}}$

So we got our answer.

$\\\;\large{\underline{\rm{Thus,\;either\;the\;value\;of\;x\;is\;\;\boxed{\bf{\dfrac{(a\;+\;b)}{2}}}\;\rm{or}\;\;\boxed{\bf{\dfrac{(a\;-\;b)}{2}}}}}}$

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2.) 9x² - 6b²x - (a⁴ - b⁴) = 0

Using identity, a⁴ - b⁴ = (a² + b²)(a² - b²), we get,

$\\\;\sf{:\rightarrow\;\;\;9x^{2}\;-\;[3(a^{2}\;+\;b^{2})\;+\;3(b^{2}\;-\;a^{2})]x\;+\;[(a^{2}\;+\;b^{2})(a^{2}\;-\;b^{2})]\;=\;\bf{0}}$

$\\\;\sf{:\rightarrow\;\;\;9x^{2}\;-3(a^{2}\;+\;b^{2})x\;+\;3(a^{2}\;-\;b^{2})x\;+\;(a^{2}\;+\;b^{2})(a^{2}\;-\;b^{2})\;=\;\bf{0}}$

Now taking 3x as common term, we get,

$\\\;\sf{:\rightarrow\;\;\;3x[3x\;-\;(a^{2}\;+\;b^{2})]\;+\;(a^{2}\;-\;b^{2})[3x\;-\;(a^{2}\;+\;b^{2})]\;=\;\bf{0}}$

Now taking the common terms as single, we get,

$\\\;\sf{:\rightarrow\;\;\;[3x\;+\;(a^{2}\;-\;b^{2})][3x\;-\;(a^{2}\;+\;b^{2})]\;=\;\bf{0}}$

Here, either

[3x + (a² - b²)] = 0, or

[3x - (a² + b²)] = 0

Then,

✒ 3x + (a² - b²) = 0 or 3x - (a² + b²) = 0

✒ 3x + a² - b² = 0 or 3x - a² - b² = 0

✒ 3x = b² - a² or 3x = a² + b²

$\\\;\quad\sf{:\mapsto\;\;x\;=\;\bf{\dfrac{(b^{2}\;-\;a^{2})}{3}}\;\;\;\sf{or\;\;\;x\;=}\;\bf{\dfrac{(a^{2}\;+\;b^{2})}{3}}}$

So we got out answer.

$\\\;\large{\underline{\rm{Thus,\;either\;the\;value\;of\;x\;is\;\;\boxed{\bf{\dfrac{(a^{2}\;+\;b^{2})}{3}}}\;\rm{or}\;\;\boxed{\bf{\dfrac{(b^{2}\;-\;a^{2})}{3}}}}}}$

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★ More Identities to know :-

$\\\;\quad\sf{\leadsto\;\;\;(a\;+\;b)^{2}\;=\;a^{2}\;+\;b^{2}\;+\;2ab}$

$\\\;\quad\sf{\leadsto\;\;\;(a\;-\;b)^{2}\;=\;a^{2}\;+\;b^{2}\;-\;2ab}$

$\\\;\quad\sf{\leadsto\;\;\;(a\;+\;b)^{2}\;-\;2ab\;=\;a^{2}\;+\;b^{2}}$

$\\\;\quad\sf{\leadsto\;\;\;(a\;+\;b\;+\;c)^{2}\;=\;a^{2}\;+\;b^{2}\;+\;c^{2}\;+\;2ab\;+\;2bc\;+\;2ac}$

$\\\;\quad\sf{\leadsto\;\;\;(a\;+\;b)^{3}\;=\;a^{3}\;+\;b^{3}\;+\;3ab(a\;+\;b)}$

$\\\;\quad\sf{\leadsto\;\;\;(a\;+\;b)[a^{2}\;-\;ab\;+\;b^{2}]\;=\;a^{3}\;+\;b^{3}}$

$\\\;\quad\sf{\leadsto\;\;\;(a\;-\;b)[a^{2}\;+\;ab\;+\;b^{2}]\;=\;a^{3}\;-\;b^{3}}$

$\\\;\quad\sf{\leadsto\;\;\;(x\;+\;a)(x\;+\;b)\;=\;x^{2}\;+\;(a\;+\;b)x\;+\;ab}$

$\\\;\quad\sf{\leadsto\;\;\;(a\;-\;b)^{3}\;=\;a^{3}\;-\;b^{3}\;-\;3ab(a\;-\;b)}$

Answer 2

Answer:

❶ 4x² - 4ax + (a² - b²) = 0

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

⇒ 2x{2x - (a + b)} - (a + b) {2x - (a + b)} = 0

⇒ {2x - (a + b)} {2x - (a - b)} = 0

⇒ 2x - (a + b) = 0 ; 2x - (a - b) = 0

⇒ 2x = (a + b) ; 2x = (a - b)

⇒ x = $\sf\dfrac{(a + b)}{2}$ ; x = $\sf\dfrac{(a - b)}{2}$

➠ x = $\sf\dfrac{(a + b)}{2}$ , $\sf\dfrac{(a - b)}{2}$

❷ 9x² - 6b²x - (a⁴ - b⁴) = 0

⇒ 9x² - 3(2b²)x - (a⁴ - b⁴) = 0

⇒ 9x² - 3(b² - a² + b² + a²)x + (b⁴ - a⁴) = 0

⇒9x²- 3(b² - a²)x - 3(b² + a²)x + (b² + a²)(b² - a²)= 0

⇒ 3x(3x - b² + a²) - (b² + a²) (3x - b² + a²) = 0

⇒ (3x - b² - a²) (3x - b² + a²) = 0

⇒ 3x - b² - a² = 0 ; 3x - b² + a² = 0

⇒ 3x = a² + b² ; 3x = b² - a²

⇒ x = $\sf\dfrac{{a}^{2} + {b}^{2}}{3}$ ; x = $\sf\dfrac{{b}^{2} - {a}^{2}}{3}$

➠ x = $\sf\dfrac{{a}^{2} + {b}^{2}}{3}$ , $\sf\dfrac{{b}^{2} -{a}^{2}}{3}$