Quadratic formula

Definition: With the quadratic formula we solve the quadratic equation: { ax^2+bx+c=0} .

The formula is: x_1,x_2 = {{ - b \,\pm\, \sqrt {b^2 \,-\, 4 \cdot a \cdot c} } \over {2 \cdot a}}

Rules

The formula can give 0, 1 or 2 real number solutions (only real numbers exist on a graph!).
If there are two different solutions they are written x_1 and x_2
The symbol \pm means that for x_1 we use the + sign and for x_2 we use the - sign.

The discriminant is: D=b^2-4\cdot a\cdot c (the expression inside the square root).

Example

Solve the equation: x^2-2x-3=0 for x.

Solution: Here а=1 \quad b=-2 \quad c=-3 (other examples ?)

x_1 ,\,x_2 =\, {{ - ( - 2) \,\pm\, \sqrt {( - 2)^2 - 4 \cdot 1 \cdot ( - 3)} } \over {2 \cdot 1}} = {{2 \,\pm\, \sqrt {4 + 12} } \over 2} = {{2 \,\pm\, 4} \over 2}

x_1 = {{2 + 4} \over 2} = {6 \over 2} = 3 and x_2 = {{2 - 4} \over 2} = {{ - 2} \over 2} = - 1

Answer: x_1=3 and x_2=-1 are the two solutions to the equation f(x)=x^2-2x-3=0 (see the left graph below!).

Rules Explanation

Depending on the function, the discriminant can be positive, zero or negative.

If the discriminant is positive, the formula gives two distinct numbers: x_1 and x_2 .

In the above example, the discriminant was 16 (positive) and so the quadratic function f(x)=x^2-2x-3 has two distinct roots. This means the graph of the function has two x-intercepts x_1=3 and x_2=-1 . See left graph below.

If the discriminant is zero, the formula give one root and the function "touches" the x-axis at this one point. See middle graph below.
If the discrimnant is negative, the formula has no real solution and since graphs are "real", the graph cannot cross or touch the x-axis. See right graph below.

The discriminant D>0	The discriminant D=0	The discriminant D<0
D>0 - the parabola crosses the x-axis in two distinct points	D=0 - the parabola touches the x-axis at a unique point	D<0 - the parabola doesn't cross or touch the x-axis

sample function: f(x)=x^2-2x-3	sample function: f(x)=x^2-4x+4	sample function: f(x)=2x^2+1
We solve: x^2-2x-3=0	We solve: x^2-4x+4=0	We solve: 2x^2+1=0
D= ( - 2)^2 - 4 \cdot 1 \cdot ( - 3)=16>0	D= ( - 4)^2 - 4 \cdot 1 \cdot 4=0	D= ( 0)^2 - 4 \cdot 2 \cdot 2=-8<0
two different roots	one unique root	no roots
x_1=3 and x_2=-1	x_1=x_2=2	x_1 and x_2 do not exist!