Question

Find the zeroes of the quadratic polynomial: x²/a+b/ac*x/b-1/c

Answer 1

Answer:

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

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