Among the ways to find the numerical value of x, a process also known as find the roots of an equation or find the solution of an equation, stand out: Bhaskara formula it's the process of completing squares. The latter is the focus of today's text.
The number of solutions to an equation is given by its degree. Therefore, first-degree equations have only one solution, third-degree equations have three solutions, and quadratic equations have two solutions, also called roots..
Second degree equations, in their reduced form, can be written as follows:
ax2 + bx + c = 0
square completion method
In which case the quadratic equation is a perfect square trinomial
Second degree equations resulting from a remarkable product are known as perfect square trinomial. To find its roots, we will use the method exemplified below:
Example: Calculate the roots of the x equation2 + 6x + 9 = 0.
Note that the coefficient b is 6 = 2·3. To write it in the form of a remarkable product, just check if c = 32, which is true, since 32 = 9 = c. In this way, we can write:
x2 + 6x + 9 = (x + 3)2 = 0
Note that a notable product is the product between two equal polynomials. In the case of this equation, we will have:
(x + 3)2 = (x + 3)(x + 3) = 0
A product is only equal to zero when one of its factors is equal to zero. Therefore, for (x + 3)(x + 3) = 0, it is necessary that (x + 3) = 0 or (x + 3) = 0. Hence the two equal results for the equation x2 + 6x + 9 = 0, which are: x = – 3 or x = – 3.
In short: to solve the x equation2 + 6x + 9 = 0, write:
x2 + 6x + 9 = 0
(x + 3)2 = 0
(x + 3)(x + 3) = 0
x = – 3 or x = – 3
In which case the quadratic equation is not a perfect square trinomial
An equation of the second in which coefficient b and coefficient c do not meet the relations established above is not a perfect square trinomial. In this case, the solving method highlighted above can be used with the addition of a few steps. Note the following example:
Example: Calculate the roots of the x equation2 + 6x – 7 = 0.
Note that this equation is not a perfect square trinomial. For it to be, we can use the following operations:
Note that b = 2·3, so in the first member the expression that should appear is x2 + 6x + 9, because in this expression b = 2·3 and c = 32.
For this "transformation", add 32 on the two members of this equation, "pass" the - 7 to the second member, perform the possible operations and observe the results:
x2 + 6x - 7 + 32 = 0 + 32
x2 + 6x + 32 = 32 + 7
x2 + 6x + 9 = 9 + 7
x2 + 6x + 9 = 16
(x + 3)2 = 16
√(x + 3)2 = √16
x + 3 = 4 or x + 3 = – 4
This last step must be split into two equations, as the root of 16 can either be 4 or – 4 (this only occurs in equations. If asked what the root of 16 is, the answer is just 4). So, it is necessary to find all possible results. Continuing:
x + 3 = 4 or x + 3 = – 4
x = 4 – 3 or x = – 4 – 3
x = 1 or x = – 7
In which case the coefficient "a" is not equal to 1
The previous cases are intended for second degree equations where the coefficient "a" is equal to 1. If the coefficient “a” is different from 1, just divide the whole equation by the value of “a” and proceed with the calculations in the same way as in the previous case.
Example: Calculate 2x roots2 + 16x – 18 = 0
Note that a = 2. So divide the entire equation by 2 and simplify the results:
2x2 + 16x – 18 = 0
2 2 2 2
x2 + 8x – 9 = 0
Once this is done, repeat the procedures of the previous case.
x2 + 8x – 9 = 0
x2 + 8x – 9 + 16 = 0 + 16
x2 + 8x + 16 = 9 + 16
(x + 4)2 = 25
√(x + 4)2 = √25
x + 4 = 5 or x + 4 = –5
x = 5 – 4 or x = – 5 – 4
x = 1 or x = – 9
Notable Products and Second Degree Equations: Origin of the Square Completion Method
The quadratic equations are much like the remarkable products sum square and square of the difference.
The sum squared, for example, is a sum of two monomials squared. Watch:
(x + k)2 = x2 + 2kx + k2
The first member of the above equality is known as remarkable product and the second how perfect square trinomial. The latter is very much like an equation of the second degree. Watch:
Perfect square trinomial: x2 + 2kx + k2
Second degree equation: ax2 + bx + c = 0
That way, if there's any way to write a quadratic equation as a remarkable product, maybe there is also a way to find your results without the need to use the formula of Bhaskara.
To do this, note that, in the notable product above, a = 1, b = 2·k and c = k2. In this way, it is possible to write equations that fulfill these requirements in the form of a remarkable product.
So look at the coefficients in the equation. If “a” is different from 1, divide the entire equation by the value of “a”. Otherwise, observe coefficient “b”. The numerical value of half of this coefficient must equal the numerical value of the square root of the coefficient “c”. Mathematically, given the equation ax2 + bx + c = 0, if a = 1 and in addition:
B = c
2
So, you can write this equation like this:
ax2 + bx + c = (x + B) = 0
2
And its roots will be - B and + b.
2 2
Hence all the theory used to calculate roots of quadratic equations by the method of completing squares.
By Luiz Paulo Moreira
Graduated in Mathematics
Source: Brazil School - https://brasilescola.uol.com.br/matematica/metodo-completar-quadrados.htm