Question

Solve: 16x2 – 8a²x + (a4 - b4) = 0 for x

class 10 quadratic equations​

Answer 1

Answer:

using formula and solving

Answer 2

The required solution is

$\boxed{ \: \: \displaystyle \bf \: x = \frac{{a}^{2} - {b}^{2} }{4}, \frac{{a}^{2} + {b}^{2} }{4} \: \: }$

Given :

The equation

$\displaystyle \sf{ 16 {x}^{2} - 8 {a}^{2} x + ( {a}^{4} - {b}^{4} ) = 0 }$

To find :

The value of x

Solution :

Step 1 of 2 :

Write down the given equation

Here the given equation is

$\displaystyle \sf{ 16 {x}^{2} - 8 {a}^{2} x + ( {a}^{4} - {b}^{4} ) = 0 }$

Step 2 of 2 :

Find the value of x

$\displaystyle \sf{ 16 {x}^{2} - 8 {a}^{2} x + ( {a}^{4} - {b}^{4} ) = 0 }$

$\displaystyle \sf{ \implies } 16 {x}^{2} - 8 {a}^{2} x + {a}^{4} - {b}^{4} = 0$

$\displaystyle \sf{ \implies } {(4x)}^{2} - 2.4x. {a}^{2} + {( {a}^{2} )}^{2} - {b}^{4} = 0$

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

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

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

Now 4x - a² + b² = 0 gives

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

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

Again 4x - a² - b² = 0 gives

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

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

Hence the required solution is

$\displaystyle \sf \: x = \frac{{a}^{2} - {b}^{2} }{4}, \frac{{a}^{2} + {b}^{2} }{4}$

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