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A quadratic equation in the standard form is given by
ax 2 + bx + c = 0
where a, b and c are constants with a not equal to zero.
Solve the above equation to find the quadratic fomulas
- Given
ax 2 + bx + c = 0
- Divide all terms by a
x 2 + (b / a) x + c / a = 0
- Subtract c / a from both sides
x 2 + (b / a) x + c / a - c / a = - c / a
- and simplify
x 2 + (b / a) x = - c / a
- Add (b / 2a) 2 to both sides
x 2 + (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 = [ b 2 - 4a c ] / ( 4a2 )
- Solve by taking the square root
x + (b / 2a) = ± sqrt { [ b 2 - 4a c ] / ( 4a2 ) }
- Solve for x to obtain two solutions
x = - b / 2a ± sqrt { [ b 2 - 4a c ] / ( 4a2 ) }
- The term sqrt { [ b 2 - 4a c ] / ( 4a2 ) } may be written
sqrt { [ b 2 - 4a c ] / ( 4a2 ) } = sqrt(b 2 - 4a c) / 2 | a |
- Since 2 | a | = 2a when a > 0 and 2 | a | = -2a when a < 0, the two solutions to the quadratic equation may be written
x = [ -b + sqrt( b 2 - 4a c ) ] / 2 a
x = [ -b - sqrt( b 2 - 4a c ) ] / 2 a
- The term b 2 - 4a c which is under the square root in both solutions is called the discriminant of the quadratic equation. It can be used to determine the number and nature of the solutions of the quadratic equation. 3 cases are possible
case 1: If b 2 - 4a c > 0 , the equation has 2 solutions.
case 2: If b 2 - 4a c = 0 , the equation has one solutions of mutliplicity 2.
case 3: If b 2 - 4a c < 0 , the equation has 2 imaginary solutions.
More references and links on how to Solve Equations, Systems of Equations and Inequalities.
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