Quadratic Formula

The General Quadratic Equation

This equation

ax2+bx+c=0ax^2 + bx +c = 0

is the general form of what’s called a second-degree or quadratic equation. A quadratic equation is any equation in which the highest power or exponent of the variable x is 2.

Examples:

x2+2x+1=0x^2 + 2x +1 = 0
3x2+4x3=03x^2 + 4x – 3 = 0
5x225x14=05x^2 – 25x – 14 = 0

In the first equation, a = 1, b = 2, and c = 1. In the second equation, a = 3, b = 4, and c = -3. In the third equation, a = 5, b = -25, and c = -14.

A quadratic equation doesn’t have to have the general form, but it can be put into that form. For example, suppose you have this equation:

5x2+3x=2-5x^2 + 3x = -2

Then by adding 2 to each side of the equation you get:

5x2+3x+2=0-5x^2 + 3x + 2 = 0

and then multiplying each side of the equation by -1, you get:

5x23x2=05x^2 – 3x -2= 0

which is in the general form, with a = 5, b = -3, and c = -2.

Quadratic Formula

Every quadratic equation can be solved by simply putting the equation into the general form

ax2+bx+c=0ax^2 + bx +c = 0

and then plugging the values for a, b, and c into this formula:

x=b±b24ac2ax = \frac { – b \pm \sqrt {b^2 – 4ac} } {{2a}}

The

±\pm

sign means, “plus or minus.” This means that there are actually two solutions defined here – one where a plus sign is placed before the result of the radical, and one where a minus sign is placed before the result. Here are some examples:

x2+3x+1=0x^2 + 3x +1 = 0

Using the quadratic formula, with a = 1, b = 3, and c = 1, we get

b24ac=94=5b^2 – 4ac = 9 – 4 = 5

so the solutions are

x=32±52x = \frac{-3}{2} \pm \frac{\sqrt{5}}{2}

which equals

x=32+52x = \frac{-3}{2} + \frac{\sqrt{5}}{2}

and

x=3252x = \frac{-3}{2} – \frac{\sqrt{5}}{2}

Second example:

x2+4x+5=0x^2 + 4x +5 = 0

In this case, a = 1, b = 4, and c = 5, so

b24ac=1620=4b^2 – 4ac = 16 – 20 = -4

and since

4=2i\sqrt{-4} = 2i

then the solutions from the quadratic formula are:

x=2±ix = -2 \pm i

Third example:

x2+4x+3=0x^2 + 4x +3 = 0

In this example,

b24ac=1612=4b^2 – 4ac = 16 – 12 = 4

so therefore the solutions are

x=2±1x = -2 \pm 1

which computes to x = -1 and x = -3.