The quadratic formula,

for the general equation ax²+bx+c=0

can be derived using the method of “completing the square” as follows.
starting with

ax²+bx+c=0

divide through by a

x²+bx/a+c/a=0

take the c part to the other side

x²+bx/a=-c/a

In order to complete the square we need to add half the coefficient of x (that’s b/2a) squared to both sides so we get

x²+bx/a + b²/4a²=-c/a +b²/4a²

Now we can factorise the left into a squared bracket (you can check this by multiplying it back out)

(x + b/2a)²=-c/a +b²/4a²

So if we square root both sides and take b/2a to the other side we have x on its own

x = -b/2a ±√(-c/a +b²/4a²)

Now to get this in the more traditional form we can take out 1/2a (Which means each term in the root is timesed by 4a²) from the square root to put the whole thing over 2a

x = (-b ±√(b²-4ac))/2a

Quadratic Solver

Fill in the boxes for the coefficients below to solve any quadratic equation (is also able to find complex roots)

Also see

• Proof of Cosine Rule (4)
• Area of a Triangle (3)
• Proof by Contradiction (0)
• Rational and Irrational Numbers (5)
• How to use Surds – Add, Multiply and Rationalise (1)
• Trigonometry Identities (11)

To perform calculations we are going to use the triangle above.
The three main relationships are:

Tan(x) = o/a
Sin(x) = o/h
Cos(x) = a/h

so if h = 5 and x = 30
a = Cos(30)h = 4.330

We can also use a inverse of the functions
ie) x = tan^-1(o/a)
x = sin^-1(o/h)
x = cos^-1(a/h)

so if o = 5 and a = 10
x = tan^-1(5/10) = 26.565

Using this information we can work out any side or angle in a right angled triangle as long as we have to other pieces of information (like a side and a angle or 2 sides). This is used a lot in resolving forces in physics and allows us to derive some other more complex equations.

Also see

• Trigonometry Identities (11)
• Sec, Cosec, Cot (0)
• Tan = sin/cos (6)
• CAST Diagarams for finding values of sin, cos and tan between 0 and 360 or 2 pi (0)
• Tan Graph – y=tan(x) (0)
• Compound Angles – sin(A+B) = cosAsinB+sinAcosB (6)