# Finding Roots of Quadratic Equations

__Finding Roots of Quadratic Equations__

__What is a root to an equation?__

A root to an equation is a set of value(s) that satisfy the equation and when shown graphically they are the x values at which the function intercepts the x-axis.

__The general form of a quadratic equation is:__

ax^{2} + bx + c = 0 where a, b and c are real coefficients

and before attempting to solve any quadratic function, you should always aim to get it into this form first.

If it is not in the correct form it can be converted by adding and subtracting each side functions of x in the initial form, for example:

x^{2 }= 2x – 12 (Subtract both sides by 2x)

x^{2 }– 2x = -12 (Add 12 to both sides)

x^{2 }– 2x + 12 = 0

As you can now see the previous equation is now in the standard form ax^{2} + bx + c = 0, where a = 1, b = 2 and c = 12.

__Finding the Root(s) of Quadratic Equations__

The first way to solve a quadratic equation is by __factorising__ it, an example is:

x^{2 }+ 7x + 12 = 0 --> (x + 3)(x + 4) = 0

the root to the equation is then given by the negative coefficient of the real number inside the bracket, hence is -3 and -4. A sketch of this graph would consist of a U shape intercepting the x-axis at -3 and -4.

The second way to solve a quadratic equation is to __complete the square__, an example is:

x^{2 }- 10x + 25 = 0 --> (x-5)^{2} = 9

The root to this equation is then worked out by square rooting each side and adding 5 to both sides, giving 8 and -2.

The final way to solve them by a __quadratic formula__, which is:

x = (-b +/- sqr(b^{2} – 4ac))/2a

The quadratic formula contains the function b^{2} – 4ac, this is called the discriminant and a, b and c are the coefficients of the equation when in the standard form. The value of the discriminant can show how many roots are present for a particular equation:

b^{2} – 4ac > 0 2 real roots

b^{2} – 4ac = 0 1 real root

b^{2} – 4ac < 0 2 imaginary roots (Complex conjugates)

__Example 1__

x^{2} + 6x + 3 = 0 a=1, b=6 and c=3

b^{2} – 4ac = 36 – 12 = 24

hence x = (-6 +/- 2sqr6)/2 = -3 +/- sqr6

so the two roots are -3 + sqr6 and -3 – sqr6

__Example 2__

x^{2} + 2x + 1 = 0 a=1, b=2 and c=1

b^{2} – 4ac = 0

hence x = -2/2 = -1

__Example 3__

x^{2} + 8x + 25 = 0 a=1, b=8 and c=25

b^{2} – 4ac = 64 – 100 = -36

The discriminant is less than 0, which shows that 2 complex conjugate roots are the solutions to the equation.

Since we can not find the square root of a negative number, we instead denote the term i, this represents the square root of -1 and also shows that i^{2} = -1. This now allows the solution to be found:

x = (-8 +/- 6i)/2 = -4 +/- 3i

hence the solutions are -4 + 3i and -4 – 3i which are complex conjugates