Higher second degrees equations


A cubic equation has its standard form as $$a x^3 + b x^2 +c x + d = 0$$ with $a \neq 0$. That means, every third degree equation we need to solve, first we have to reduce it to the standard form. For, if we have to solve $$ x^2 + \frac{3}{x} - 5x +1 =0$$ first we need to evaluate the least common denominator and transform it to the equivalent equation $$ x^3 - 5 x^2 + x + 3 =0$$

In the above cubic equation $a = 1, b = -5, c=1, d=3$. (Note that, here, $x$ must not be zero, otherwise the equation has no sense!)

Now, to start using some methods of solving cubics, we have to check whether the equation has a $d$ value, that means a constant (we denoted this constant by $d$). 

  1. The cubic has the $d$ constant

In this case, we can easily use Ruffini's rule for solving cubic equations. First, we need to find all numbers (positive and negative numbers) that divide the $d$ constant. Referring to the above example, since $d = 3$, all required numbers are: $$\pm 1, \pm 3$$

Then, we have to evaluate the cubic equation for the chosen numbers till when it becomes zero. When we find the number that transforms the cubic equation to zero, we can stop our search. 

For, using the above example, we have to evaluate $p(-1), p(1), p(-3)$ and $p(3)$ where $$p(x) = x^3 - 5 x^2 + x + 3 $$ Let us start by $x=-1, +1, ....$.

$$p(-1) = (-1)^3 - 5 \cdot (-1)^2 + (-1) + 3 = -1 -5 -1 +3 = -4 \neq 0$$

$$p(1) = 1^3-5\cdot 1^2 +1+3 = 1 - 5 + 1 + 3 = 0 $$

We have already found the divisor of $d$ that makes zero the cubic equation, so we can stop the above procedure. Now, we have to draw the following table, where we first place the just found number on the bottom left. Then, we have to write in the first line, all numbers that multiply the decreasing power of $x$ even if a power of $x$ is missing (using the given example that means: $1, -5, 1, 3$). Remind that the $d$ number has to be put on the right top, far from the other numbers, to the other side of the second vertical line of the table (it should be clear looking the following table):



Then, we have to put down the first number in the first line to the bottom line and multiply it for the chosen divisor. Its result has to be written in the second column and second line (just down $-5$) and it has to be added to the above number. The result of the sum ($-5 + 1 = -4$) has to be written down the horizontal line of the table. The procedure is repeated till we find zero as final result. 

Now, $x =1$ is for sure a solution of the cubic equation. We have to find, if they exist, other two solutions. They come from the solution of a second degree equation and its coefficient are given by the last line of the table. 

$$1 x^2 - 4 x - 3=0$$ Solving it, we find $2 \pm \sqrt{7}$ as solutions. 

NOTE: if $a \neq 1$, the numbers that make zero the equation have to be searched from all the fractions: $$\frac{\mbox{divisors of} \quad d}{\mbox{divisors of} \quad a}$$. 


2. The cubic does NOT have the $d$ constant

If the cubic has the form $$ax^3 + bx^2+cx=0$$ where the $d$ constant is missing, we have to factor an $x$ out of the equation. Indeed, since the equation does not have the constant, every term of the equation has an $x$ as a factor. We can simplify the equation factoring the $x$ out of the equation and re-writing it as follows: $$x \cdot (ax^2 + bx + c) =0$$

Now, we have to apply the Null Factor Law to find the roots of the equation. That means, we have to solve

$$x=0 \quad vel \quad  ax^2 + bx +c =0$$

For, if the starting cubic equation is $$x^3+x^2-x =0$$ we factor the $x$ out of the equation and we find $$x \cdot ( x^2 + x -1) =0$$ which solutions are $$x=0 \quad \mbox{or} \quad  x_{1/2} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot (-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$$



biquadratic equation is an equation where all odd powers of $x$ are missing. That means, it will appear even-degree terms only. We will treat the simplest case, where a biquadratic equation has the form $$ax^4 + bx^2 + c =0$$ To solve biquadratic equations we have to change $$x^2 = t \quad x^4 = t^2$$ so that the equation takes the form of a quadratic equation with $t$ as variable $$at^2 + bt + c =0$$

For every positive value of $t$ we will find two values of $x= \pm \sqrt{t}.$

Let us consider the biquadratic equation $$3x^4 + 2x^2 - 1 = 0$$ Then, we change $$x^2 = t \quad x^4 = t^2$$ so that the equation takes the form $$3 t^2 + 2 t - 1 =0$$ Its solutions are $$t_{1/2} = \frac{-1 \pm \sqrt{1^2 - 3 \cdot  (-1)}}{3} = \frac{-1 \pm \sqrt{4}}{3} = \frac{-1 \pm 2}{3}$$

$$t_1 = \frac{1}{3} \Longrightarrow x = \pm \frac{1}{\sqrt{3}} = \pm \frac{\sqrt{3}}{3}$$ and 

$$t_2 = \frac{-3}{3} = -1  \Longrightarrow \mbox{there are no other values of } x \quad  \mbox{since}$$

$$x = \pm \sqrt{-1} \mbox{ is impossible because the base of the root is negative.}$$



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