One of the techniques used to solve quadratic equations is the method known as complete squares. This method consists of interpreting the equation of seconddegree as a perfect square trinomial and write your factored form. Sometimes this simple procedure already reveals the roots of the equation.
Therefore, it is necessary to have basic knowledge about notable products, trinomialsquarePerfect and polynomial factorization to use this technique. Often, however, it allows calculations to be done “in the head”.
Therefore, we will recall the three cases of productsremarkable before demonstrating the methodto completesquares, which, in turn, will be exposed in three different cases.
Outstanding products and perfect square trinomials
Next, see the remarkable product, the trinomialsquarePerfect which is equivalent to it and the shape factored of this trinomial, respectively. To do so, consider that x is unknown and The is any real number.
(x + k)2 = x2 + 2kx + k2 = (x + k)(x + k)
(x - k)2 = x2 – 2kx + k2 = (x - k)(x - k)
The equation of the second degree referring to the third productremarkable, known as the product of the sum and the difference, can be solved using a technique that makes calculations even easier. As a result, it will not be considered here.
The equation is the perfect square trinomial
If one equation of seconddegree is a perfect square trinomial, then you can identify its coefficients as: a = 1, b = 2k or – 2k and c = k2. To check this, just compare a quadratic equation with a trinomialsquarePerfect.
Therefore, in the solution of the equation of seconddegree x2 + 2kx + k2 = 0, we will always have the possibility to do:
x2 + 2kx + k2 = 0
(x + k)2 = 0
√[(x + k)2] = √0
|x + k| = 0
x + k = 0
x = - k
– x – k = 0
x = - k
Thus, the solution is unique and equal to –k.
If equation be x2 – 2kx + k2 = 0, we can do the same:
x2 – 2kx + k2 = 0
(x - k)2 = 0
√[(x - k)2] = √0
|x – k| = 0
x - k = 0
x = k
– x + k = 0
– x = – k
x = k
Therefore, the solution is unique and equal to k.
Example: What are the roots of equation x2 + 16x + 64 = 0?
Note that the equation is a trinomialsquarePerfect, since 2k = 16, where k = 8, and k2 = 64, where k = 8. So we can write:
x2 + 16x + 64 = 0
(x + 8)2 = 0
√[(x + 8)2] = √0
x + 8 = 0
x = – 8
Here the result has been simplified, as we already know that the two solutions will be equal to the same real number.
The equation is not a perfect square trinomial
In cases where the equation of seconddegree is not a perfect square trinomial, we can consider the following hypothesis to calculate its results:
x2 + 2kx + C = 0
Note that for this equation to turn into a trinomialsquarePerfect, just replace the value of C with the value of k2. Since this is an equation, the only way to do this is to add k2 on both members, then swapping the member coefficient C. Watch:
x2 + 2kx + C = 0
x2 + 2kx + C + k2 = 0 + k2
x2 + 2kx + k2 = k2 - Ç
After this procedure, we can proceed with the previous technique, transforming the trinomialsquarePerfect into remarkable product and calculating the square roots on both limbs.
x2 + 2kx + k2 = k2 - Ç
(x + k)2 = k2 - Ç
√[(x + k)2] = √(k2 - Ç)
x + k = ± √(k2 - Ç)
The ± sign appears whenever the result of a equation is a square root, because in these cases the square root result is a module, as shown in the first example. Finally, all that's left is to do:
x = – k ± √(k2 - Ç)
So, these equations have two results real and distinct, or no real result when C > k2.
For example, calculate the roots of x2 + 6x + 8 = 0.
Solution: Note that 6 = 2·3x. Hence, k = 3 and therefore k2 = 9. Therefore, the number we must add in both members is equal to 9:
x2 + 6x + 8 = 0
x2 + 6x + 8 + 9 = 0 + 9
x2 + 6x + 9 = 9 - 8
x2 + 6x + 9 = 1
(x + 3)2 = 1
√[(x + 3)2] = ± √1
x + 3 = ± 1
x = ± 1 - 3
x’ = 1 – 3 = – 2
x’’ = – 1 – 3 = – 4
In which case the coefficient a ≠ 1
when the coefficient The, gives equation of seconddegree, is different from 1, just divide the whole equation by the numerical value of the coefficient The to then apply one of the two previous methods.
So, in the 2x equation2 + 32x + 128 = 0, we have the unique root equal to 8, because:
2x2+ 32x + 128 = 0
2 2 2 2
x2 + 16x + 64 = 0
And, in the 3x equation2 + 18x + 24 = 0, we have the roots – 2 and – 4, because:
3x2 + 18x + 24 = 0
3 3 3 3
x2 + 6x + 8 = 0
By Luiz Paulo Moreira
Graduated in Mathematics
Source: Brazil School - https://brasilescola.uol.com.br/o-que-e/matematica/o-que-e-metodo-completar-quadrados.htm
