Many quadratic equations can be factorized to find the roots of the equation. Others have roots that are either fractional, irrational, or complex. In many cases, the general equation for finding the roots needs to be used. This formula is derived here.
Quadratic Equation Solution Formula
For the general quadratic equation ax² + bx + c = 0, the roots may be found by substituting into the formula:
x = (-b± √(b² - 4.a.c)) / (2.a)
For example, the solution to the quadratic equation
2x² - 3.x -2
is
x =[ -(-3) ± √[ (-3)² - 4.(2).(-2)] / (2.2) ]
(Note that the period is used to show multiplication, not decimal point, so 2.2 = 2×2 = 4)
This simplifies to
[3 ± √ (9 + 16) ] / 4
= ( 3 ± √ 25 ) / 4
= (3 ± 5) / 4
= 8 / 4 or -2 / 4
i.e. x = 2 or -½
While this particular example could have been factorized easily enough, it allows the solution to be verified.
Quadratic Equation Solution Proof
ax² + bx + c = 0 .... equation [1]
Dividing throughout by a:
x² + b.x/a + c/a = 0 ... [equation 2]
Now, for all values of x and y, (x + y)² = x² + 2.x.y + y² [equation 3]
So comparing equations 2 and 3, the x² term is common, but to get the x-terms to be equivalent:
b / a = 2.y
i.e. y = b / (2.a)
Now, (x + b / (2a) )² = x² + 2.b.x / (2a) + (b / 2.a)² ... [equation 4]
= x² + b.x / (a) + (b / 2.a)²
The left hand side of equation 2 is
x² + b.x/a + c/a, but from equation 4 it is known that
(x + b / (2a) )² = x² + b.x / (a) + (b / 2.a)²
Now, and this is the key manipulation,
x² + b.x/a + c/a = x² + b.x / (a) + (b / 2.a)² - (b / 2.a)² + c / a
= (x + b / (2a))² - (b / 2.a)² + c / a
The goal is to set this equal to zero (to find the roots of the equation), so
(x + b / (2a))² - (b / 2.a)² + c / a = 0
So
(x + b / (2.a))² = (b / 2.a)² - c / a
(x + b / (2.a))² = b² / 4.a² - c / a
(x + b / (2.a))² = b² / 4.a² - 4.a.c / 4.a²
Now take the square root of each side:
x + b / (2.a) = √(b² - 4.a.c) / (2.a)
So
x = -b/2.a ± √(b² - 4.a.c) / (2.a)
=(-b± √(b² - 4.a.c)) / (2.a)
Q.E.D.
Quadratic Equation Solution Summary
The formula to solve quadratic equations is used very extensively. An example of how to use it is shown here, as well as the general proof for the formula. The formula is best used when factorisation of the quadratic equation is not easily done, or when the roots of the equation are complex.
Quadratic Equation Solution References
Since the equation is derived from first principles, no references are really needed. There are many excellent mathematics texts available, especially those that deal with algebraic manipulation.
Martin Bell - Martin holds a B.Sc. degree in chemical engineering, and an M.Sc. degree in electronics and computing. He has spent more than 25 years ...
Join the Conversation