what are the steps to solve this ax2 + bx + c = 0"

Question

how do we solve it

 

Answer 1

This answer comes from Littlewood's A University Algebra

ax²+bx+c=0  Multiply by 4a to get

4a²x²+4abx+4ac=0 Compete the square as

(2ax+b)²=4a²x²+4abx+b² so that

 

4a²x²+4abx+4ac=(2ax+b)²-b²+4ac=0

(2ax+b)²=b²-4ac

2ax+b=±√(b²-4ac)

2ax=-b±√(b²-4ac)

x=(-b±√(b²-4ac))/(2a)

Answer 2

Quadratic equation is derived by completing the square:

 

ax^2 + bx + c = 0

 

Subtract c from both sides:

ax^2 + bx = -c

 

Divide by a:

x^2 + (b/a)x = -c/a

 

Complete the square by taking half of b/a  and squaring it, then adding to both sides:

Note: (b/a) * 1/2 = b/(2a)

x^2 + (b/a)x + (b/(2a))^2 = -c/a + (b/(2a))^2

 

Simplify:

 

x^2 + (b/a)x + b^2/(4a^2) = -c/a + b^2/(4a^2)

 

Factor:

(x - (b/(2a))))^2 = -c/a + b^2/(4a^2)

 

Take the square root:

 

x + (b/(2a)) = +/-  sqrt(-c/a + b^2/(4a^2))

 

Subtract b/(2a) from both sides:

 

x = -b/(2a) +/- sqrt(-c/a + b^2/(4a^2))

 

Simplify:

 

x = -b/(2a) +/- sqrt(  -4ac             b^2          )

                               -------  +   ---------

                                4a^2           4a^2

 

 

x = -b +/- sqrt (b^2 - 4ac)

      --------------------------

                 2a