Question

solve for x: 4x^2-4a^2x+(a^4-b^4)=0

Answer 1

Here we go buddy,

$4 {x}^{2} - 4 {a}^{2} x + ( {a}^{4} - {b}^{4} ) \\ \\ 4 {x}^{2} - 4 {a}^{2} x + ( {a}^{2} - {b}^{2} )( {a}^{2} + {b}^{2} ) \\ \\ 4 {x}^{2} - 2( {a}^{2} - {b}^{2} )x - 2( {a}^{2} + {b}^{2} ) x+ ( {a}^{2} - {b}^{2} )( {a}^{2} + {b}^{2} ) \\ \\ 2x(2x - ( {a}^{2} - {b}^{2} )) - ( {a}^{2} + {b}^{2})(2x - ( {a}^{2} - {b}^{2} )) \\ \\ (2x - ( {a}^{2} + {b}^{2} ))(2x - ( {a}^{2} - {b}^{2} ))$
From here,

$2x - ( {a}^{2} + {b}^{2} ) = 0 \\ 2x = ( {a}^{2} + {b}^{2}) \\ x = \frac{( {a}^{2} + {b}^{2} )}{2}$

Or

$2x - ( {a}^{2} - {b}^{2} ) = 0 \\ 2x = ( {a}^{2} - {b}^{2}) \\ x = \frac{( {a}^{2} - {b}^{2} )}{2}$