Question

solve the quadratic equation for x. 4x²-4a²x+(a⁴-b⁴)=0
solve this and get 50 pts

Answer 1

4x2 - 4a2x +a4 = b4

=>( 2x -a2 )2  =  (b2)2

=> 2x - a2 = + b2 or  - b2

=> x = (a2 - b2 ) / 2  or ( a2  + b2 ) /2

Answer 2

i) 4x² - 4a²x + a⁴ - b⁴ = 0

ii) Grouping, {(2x)² - 2(2x)(a²) + (a²)²} - (b²)² = 0

==> (2x - a²)² - (b²)² = 0

iii) The above is of the form a² - b² = (a - b)(a + b),

here a = (2x - a²) and b = b²

So, (2x - a²)² - (b²)² = {(2x - a²) + (b²)}*{(2x - a²) - (b²)}

= {2x - (a² - b²)}*{2x - (a² + b²)} = 0

==> Either {2x - (a² - b²)} = 0 or {2x - (a² + b²)} = 0

So, when {2x - (a² - b²)} = 0, x = (a² - b²)/2

and when {2x - (a² + b²)} = 0, x = (a² + b²)/2