Question

4 x square - 4 a square x + a to the power 4 minus b to the power 4 is equal to zero solve

Answer 1

Tip:

Recall the concept of algebraic identities

$(a-b)^2=a^2-2ab+b^2$

$a^2-b^2=(a+b)(a-b)$

Given: $4x^2-4a^2x+a^4-b^4=0$

Explanation:

The equation is $4x^2-4a^2x+a^4-b^4=0$

Using the identity, $(a-b)^2=a^2-2ab+b^2$, the equation can be written as

$\implies (2x-a^2)^2-b^4=0$

This can be written as

$\implies (2x-a^2)^2-(b^2)^2=0$

Using the identity $a^2-b^2=(a+b)(a-b)$

$\implies (2x-a^2+b^2)(2x-a^2-b^2)=0$

$\implies (2x-a^2+b^2)=0,(2x-a^2-b^2)=0$

When, $(2x-a^2+b^2)=0,$

$\implies 2x=a^2-b^2$

$\implies x=\frac {a^2-b^2}{2}$

When, $\implies (2x-a^2-b^2)=0,$

$\implies 2x=a^2+b^2$

$\implies x=\frac {a^2+b^2}{2}$

Hence, the value of $x=\frac {a^2-b^2}{2}, x=\frac {a^2+b^2}{2}$

Answer 2

The required solution is

$\boxed{ \: \: \displaystyle \bf \: x = \frac{{a}^{2} - {b}^{2} }{2}, \frac{{a}^{2} + {b}^{2} }{2} \: \: }$

Given :

The equation

$\displaystyle \sf{ 4 {x}^{2} - 4 {a}^{2} x + ( {a}^{4} - {b}^{4} ) = 0 }$

To find :

The value of x

Solution :

Step 1 of 2 :

Write down the given equation

Here the given equation is

$\displaystyle \sf{ 4 {x}^{2} - 4 {a}^{2} x + ( {a}^{4} - {b}^{4} ) = 0 }$

Step 2 of 2 :

Find the value of x

$\displaystyle \sf{ 4 {x}^{2} - 4 {a}^{2} x + ( {a}^{4} - {b}^{4} ) = 0 }$

$\displaystyle \sf{ \implies } 4 {x}^{2} - 4 {a}^{2} x + {a}^{4} - {b}^{4} = 0$

$\displaystyle \sf{ \implies } {(2x)}^{2} - 2.2x. {a}^{2} + {( {a}^{2} )}^{2} - {b}^{4} = 0$

$\displaystyle \sf{ \implies } {(2x - {a}^{2} )}^{2}- {b}^{4} = 0$

$\displaystyle \sf{ \implies } {(2x - {a}^{2} )}^{2}- {( {b}^{2} )}^{2} = 0$

$\displaystyle \sf{ \implies } (2x - {a}^{2} + {b}^{2} )(4x - {a}^{2} - {b}^{2} ) = 0$

Now 2x - a² + b² = 0 gives

$\displaystyle \sf{2x = {a}^{2} - {b}^{2} }$

$\displaystyle \sf{ \implies x = \frac{{a}^{2} - {b}^{2} }{2} }$

Again 2x - a² - b² = 0 gives

$\displaystyle \sf{2x = {a}^{2} + {b}^{2} }$

$\displaystyle \sf{ \implies x = \frac{{a}^{2} + {b}^{2} }{2} }$

Hence the required solution is

$\displaystyle \sf \: x = \frac{{a}^{2} - {b}^{2} }{2}, \frac{{a}^{2} + {b}^{2} }{2}$

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