ax2+bx+c=0 prove it and prove the quadratic formula
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ax2 + bx + c = 0
ax2 + bx = –c
Divide through by whatever is multiplied on the squared term.
x2 + (b / a) x = - c / a
and simplify
x2 + (b / a) x = - c / a
Add (b / 2a)2 to both sides
x2 + (b / a) x + (b / 2a)2 = - c / a+ (b / 2a)2
to complete the square
(x + b / 2a )2 = - c / a + (b / 2a)2
Group the two terms on the right side of the equation
(x + b / 2a)2 = (b2 - 4 a c) / (4 a2 )
Solve by taking the square root
x + b / 2a = ± √{ (b2 - 4 a c ) / (4 a2) }
Solve for x to obtain two solutions
x = - b / 2a ± √{ (b2 - 4 a c ) / (4 a2) }
The term √{ (b2 - 4 a c ) / (4 a2) }may be simplified as follows
√{ (b2 - 4 a c ) / (4 a2) } = √(b2 - 4 a c) / (2 |a|)
Since 2 | a | = 2 a for a > 0 and 2 | a | = -2 a for a < 0, the two solutions to the quadratic equation may be written
x = (-b + √( b2 - 4 a c)) / (2 a)
x = (-b - √ ( b2 - 4 a c)) / (2 a)
ax2 + bx = –c
Divide through by whatever is multiplied on the squared term.
x2 + (b / a) x = - c / a
and simplify
x2 + (b / a) x = - c / a
Add (b / 2a)2 to both sides
x2 + (b / a) x + (b / 2a)2 = - c / a+ (b / 2a)2
to complete the square
(x + b / 2a )2 = - c / a + (b / 2a)2
Group the two terms on the right side of the equation
(x + b / 2a)2 = (b2 - 4 a c) / (4 a2 )
Solve by taking the square root
x + b / 2a = ± √{ (b2 - 4 a c ) / (4 a2) }
Solve for x to obtain two solutions
x = - b / 2a ± √{ (b2 - 4 a c ) / (4 a2) }
The term √{ (b2 - 4 a c ) / (4 a2) }may be simplified as follows
√{ (b2 - 4 a c ) / (4 a2) } = √(b2 - 4 a c) / (2 |a|)
Since 2 | a | = 2 a for a > 0 and 2 | a | = -2 a for a < 0, the two solutions to the quadratic equation may be written
x = (-b + √( b2 - 4 a c)) / (2 a)
x = (-b - √ ( b2 - 4 a c)) / (2 a)
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