Lesson Solving quadratic equations by the quadratic formula in graphic form.

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THE QUADRATIC FORMULA IN GRAPHIC FORM

Given a quadratic function in standard form y = f(x) = ax^2 + bx + c. The graph of f(x) is a parabola that may intercept the x-axis at 2 points, or at unique point, or no point at all. That means the quadratic equation f(x) = ax^2 + bx + c = 0 may have 2 real roots, one double real root, or no real roots at all.

The quadratic formula in graphic form gives the 2 real roots by this formula:

x1 = -b/2a + d/2a  ;  and x2 = -b/2a - d/2a. (1).

In this formula:
The quantity -b/2a represents the x-coordinate of the parabola axis.
The quantities (d/2a) and (-d/2a) represents the distances from this axis to the x-intercepts of the parabola.

The quantity d may be zero, real or radical number, or imaginary number.
If d = 0: there is a double real root at x = -b/2a.
If d is a number (real or radical), there are 2 x-intercepts (or real roots)
If d is an imaginary number, there are no intercepts or no real roots.

The unknown quantity d can be computed from this relation:

d^2 = b^2 - 4ac. (2).

You can easily find the relation (2) by writing that the product of the 2 real roots is equal to c/a:

(-b/2a + d/2a)(-b/2a - d/2a) = c/a
(b^2 - d^2)/4.a^2 = c/a ---> b^2 - d^2 = 4ac ---> d^2 = b^2 - 4ac.

To solve a quadratic equation, first find d by the relation (2), then find the 2 real roots by the formula (1).

This formula (1) is simpler and easier to remember than the classic formula, since students can relate it to the parabola graph of the function f(x). The quantities (d/2a) and (-d/2a) make more sense about distances than the classic quantity "square root of b^2 - 4ac".

EXAMPLES OF SOLVING QUADRATIC EQUATIONS BY THE QUADRATIC FORMULA IN GRAPHIC FORM.

Example 1. Solve f(x) = 5x^2 + 9x - 2 = 0.
Solution. First, find d. d^2 = 81 + 4.5.2 = 121 = 11^2
We get d = 11 and d = -11. Find the 2 real roots by using formula (1);
x1 = (-9 - 11)/10 = -20/10 = -2  ;  and x2 = (-9 + 11)/10 = 2/10 = 1/5.
The 2 real roots are -2 and 1/5.

Example 2. Solve: f(x) = 6x^2 + x - 12 = 0.
Solution. First, find d. d^2 = 1 + 289 = (17)^2.
Next find the real roots by formula (1).
x1 = (-1 - 17)/12 = -18/12 = -3/2  ;  x2 = (-1 + 17)/12 = 16/12 = 4/3.
The 2 real roots are -3/2 and 4/3.

Example 3. Solve 3x^2 - 9x + 1 = 0.
Solution. d^2 = 81 - 12 = 69 ---> d = -8.31 and d = 8.31.
x1 = (9 + 8.31)/6 = 2.89  ;  and x2 = (9 - 8.31)/6 = 0.12

Example 4. Solve 9x^2 - 12x + 7 = 0.
Solution. d^2 = 144 - 252 < 0. Since d is an imaginary number, there are no real roots.