Solve by completing the square please : ax^2 + bx + c = 0.?

Answer 1

ax²+bx+c=0
Divide by a:
x² + bx/a + c/a = 0
Subtract c/a:
x² + bx/a = -c/a
Add (b/(2a))²:
x² + bx/a + b²/(4a²) = -c/a + b²/(4a²)
Factor and simplify:
(x+b/(2a))² = (b²-4ac)/(4a²)
Take the square root:
x+b/(2a) = ±√(b²-4ac)/(2a)
Subtract b/(2a):
x = (-b±√(b²-4ac))/(2a)

Answer 2

ax^2 + bx + c = 0 =>
a( x^2 + (b/a)x + (c/a) )= 0. Now assuming a is not 0 , divide by a on both sides to get
x^2 + (b/a)x + (c/a) = 0. Now complete the square.
x^2 + (b/a)x + (b/2a)^2 - (b/2a)^2 + (c/a) = 0 =>
( x + (b/2a) )^2 + ( (4ac - b^2) / 4a^2 ) = 0 =>
( x + (b/2a) )^2 ) = (b^2 - 4ac) / 4a^2. squaring both yields
x + b/2a = (+- sqrt(b^2 - 4ac)) / 2a =>
x = (- b +- sqrt(b^2 - 4ac)) / 2a

Answer 3

Usually when you have to complete the square on something, it has no leading coefficient so you'll have to get rid of that in this case.
Divide each term by a: x^2 + (b/a)x + (c/a)
Take the constant term and move it to the other side of the equation:
x^2 + (b/a)x = -(c/a)
Take half the linear term's coefficient and square it. Add that amount to the other side:
x^2 + (b/a)x + (b/2a)^2 = -(c/a) + (b/2a)^2
Now you have a perfect square:
(x + (b/2a))(x + (b/2a)) = -(c/a) + (b/2a)^2
So,
(x + (b/2a))^2 = -(c/a) + (b/2a)^2
Take the square root of both sides:
(x + (b/2a)) = Sqrt[-(c/a) + (b/2a)^2]
Subtract the constant:
x = Sqrt[-(c/a) + (b/2a)^2] -(b/2a)

Answer 4

ax^2+bx+c=0

x^2+bx/a+c/a=0
x^2+bx/a= -c/a
(x+b/2a)^2 - b^2/4a^2 = -c/a
(x+b/2a)^2 = b^2/4a^2 - c/a
(x+b/2a)^2 = (b^2 -4ac)/4a^2
x+b/2a = +and -(sqrt of b^2-4ac) /2a
x= -b/2a + and - (sqrt of b^2-4ac)/2a
x =( -b + and - (sqrt of b^2-4ac))/2a

Answer 5

Delta= b^2 - 4ac
x1= (-b+sqrt(b^2 - 4ac))/2a , x2= (-b -sqrt(b^2 - 4ac))/2a