Question

plz define the Derivation of quadratic formula in hard way
$x =(b {}^{2} + \div 4) \sqrt[2]{b { }^{2} - 4ac} \div 2a$


​

Answer 1

Step-by-step explanation:

Quadratic(2 degree) equations are written in form of ax² + bx + c = 0, where a ≠ 0. Note that if a = 0, then it is bx + c = 0 which is not quadratic. So ax² + bx + c = 0 is quadratic when a ≠ 0. On this basis,

=> ax² + bx + c = 0

Completing square:

=> a(x² + bx/a + c/a) = 0

=> x² + x(b/a) + (c/a) = 0

=> x² + 2x(b/2a) + (c/a) = 0

=> x² + 2x(b/2a) + (b/2a)² + (c/a) = (b/2a)²

=> (x + b/2a)² + (c/a) = b²/4a²

=> (x + b/2a)² = b²/4a² - c/a

=> (x + b/2a)² = (b² - 4ac)/4a²

=> x + b/2a = ± √{(b² - 4ac)/4a²}

=> x + b/2a = ± √(b² - 4ac)/2a

=> x = - b/2a ± √(b² - 4ac)/2a

=> x = {-b ± √(b² - 4ac)} / 2a