Equations of second degree.


We will do a geometric proof with an example.
What is the solution to the following equation:
x^2 + 6x - 40 = 0?


We move the independent term to the other side: x^2 + 6x = 40.
x^2 is the area of a square with side x. 6x is the area of a rectangle with sides x and 6. We can see this square and rectangle in the following figure:

With the rectangle of the figure, we construct two rectangles equals, with sides x and , and we move one of these rectangles under of the square of the figure:

Now, we also have a square with area 9, down in the fiugure, in the right corner. Finally, we obtain with all these figures, one square with side x+3.
When we have added the square with area 9, we have changed our initial equation, and we have that add 9 to both sides of the equation: x^2 + 6x + 9 = 40 + 9. Now, x^2 + 6x + 9 = (x+3)^2 = 49 and (x+3)^2 is a square with side x+3.

In both sides of equation we do the square root:

x + 3 = 7 doubleright x = 4
x+3 = -7 doubleright x=-10

The solution x=-10 is impossible in geometric, because it is negative, but it is solution of our equation.


You try to solve the following equation: x^2 + 6x - 100 = 0 as in the previous example. You can to solve the equation by using the following application. To you use the application, you can to move the point b, this point is in the segment that it is up on the left in the application. you must move the point to the figure to be right.

If you can not to see the application you can to download JAVA in the next web site: http://www.houspain.com/gttp/salsaj

General case.

We are doing the same steps of the previous example, but now we are doing for any equation of second degree: ax^2+ bx + c = 0, and we will obtain the solution: ax = {-b pm sqrt { b^2-4ac }}/{2 a}.
We have the following figure to solve the equation: x^2+ bx + c = 0:

We want to get a square with side ax, for this we multiply both sides of our equation per a. Now the calculations are easier and we have the equation: a^2x^2 + abx = -ca.

a^2x^2 is the area of the square with side ax, and bx is the area of the rectangle with sides ax and b. With this rectangle, we construct two rectangles with sides ax and b/2 as in the following figure:

We add the square with area b^2/4 (it is down in the figure, in the right corner). Now we have a square with side ax + b/2. When we have added this square to the figure, we have changed our initial equation, and we have that add b^2/4 to both sides of the equation: a^2x^2 + abx + b^2/4 = -ac + b^2/4. We multiply both sides of the last equation per 4: 4a^2x^2 + 4bxa + b^2 = -4ac + b^2, and now the calculations are easier.

4a^2x^2 + 4bxa + b^2 = (2ax + b )^2 is the area of a square with side 2ax + b. Our equation is: (2ax + b )^2 = -4ac + b^2.

We do the square root in both sides of the last equation, and we obtain:

2ax + b = sqrt{b^2-4ac}

2ax + b = -sqrt{b^2-4ac}

We work out the value of x in the last two equations, and the solutions of our equation are:

x = {-b+sqrt{b^2-4ac}}/{2 a}
x = {-b-sqrt{b^2-4ac}}/{2 a}

We will represent to b^2 – 4ac with the letter: ∆ = b^2 – 4ac. We can to have several solutions depending of the different values of∆:

1.- If ∆ > 0, the equationax^2+ bx + c = 0 have two real solutionts differents:
x = {-b+sqrt{b^2-4ac}}/{2 a}
x = {-b-sqrt{b^2-4ac}}/{2 a}

2.- If ∆ = 0, the equation ax^2+ bx + c = 0 have a double solution:
x = {-b}/{2 a}

3.- If ∆ < 0, the equation ax^2+ bx + c = 0 have not real solutions.


ecuacionesingles.txt · Última modificación: 24/04/2017 13:13 (editor externo)
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