# 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!

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Page last modified on June 06, 2010, at 08:00 AM