Question

4x²-4a²x+(a⁴-b⁴)=0
solve this question by quadratic formula {D=b²-4ac} & {x=[-b(+-)√D]/(2a)}.........
please solve this question.......​

Answer 1

Answer:

Step-by-step explanation:

4x² - 4a²x + (a⁴ - b⁴) = 0

a = 4, b = - 4a², c = ($a^{4}$ - $b^{4}$)

D = (- 4a²)² - 4×4($a^{4}$ - $b^{4}$) = 16

$x_{12}$ = (4a² ± √(16b⁴)) / 8

$x_{1}$ = (4a² + 4b²) / 8 = (a² + b²)/2

$x_{2}$ = (4a² - 4b²) / 8 = (a² - b²)/2= $\frac{1}{2}$(a - b)(a + b)  

Answer 2

Answer:

4x² - 4a²x + (a⁴ - b⁴) = 0

a = 4, b = - 4a², c = (a^{4}a4 - b^{4}b4 )

D = (- 4a²)² - 4×4(a^{4}a4 - b^{4}b4 ) = 16

x_{12}x12 = (4a² ± √(16b⁴)) / 8

x_{1}x1 = (4a² + 4b²) / 8 = (a² + b²)/2

x_{2}x2 = (4a² - 4b²) / 8 = (a² - b²)/2= \frac{1}{2}21 (a - b)(a + b)  

Step-by-step explanation:

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