Question

find the zeros of the quadratic equation
$f(x) = \frac{x {}^{2} }{a} + \frac{b}{ac} x + \frac{x}{b} - \frac{1}{c}$
pls explain the full answer
correct explanation i will mark the brainliest​

Answer 1

x²/a + (b/ac)*x + x/b - 1/c = 0

x²/a + x(b/ac + 1/b) - 1/c = 0

=> x²/a + x((b²+ac) /abc) -1/c = 0

=> bc(x²) + x(b²+ac) -ab = 0

=>x = -(b²+ac) +√(b^4 + 2ab²c +a²c² - 4ab²c) /2bc

=> x = -(b²+ac) + √(b²-ac) ² / 2bc

=> x= [ -(b²+ac) + (b²-ac) ]/2bc

=> x = [ -b²+b²-ac-ac ] / 2bc

=> x = -2ac/2bc

=> x = -a/b

=> x = [ -b² +ac -b² +ac]/2bc

=> x =[ -2b²+2ac ]/2bc

=> x =(-b²+ac) /bc

=> x = -b/c + a/b

two zeros x = -a/b , -b/c+a/b