# Getting proof: Bhaskara formula

Fonte: https://vbfelix.github.io/posts/0019-quadratic-equation/index.html

```{r setup,echo=FALSE,message=FALSE,warning=FALSE}
suppressWarnings(library(ggplot2))
suppressWarnings(library(relper))
suppressWarnings(library(dplyr))
suppressWarnings(library(tidyr))
suppressWarnings(library(janitor))
suppressWarnings(library(knitr))
suppressWarnings(library(kableExtra))
suppressWarnings(library(forcats))
```

In this post, we explore the root of the quadratic equation, i.e., Bhaskara formula.

# Context

The quadratic equation is given by:

$$
ax^2 + bx + c,
$$ {#eq-quadratic}

where:

-   $x$ is the variable;

-   $a,b,c$ are the coefficients.

Here an example of a quadratic function:

```{r, echo = FALSE}
base <-
  ggplot() +
  plt_theme_x()+
  plt_water_mark(vfx_watermark)+
  labs(y = "")+
  scale_x_continuous(
    breaks = seq(-10,10,2),
    expand = c(.01,0),
    limits = c(-10,10)
  )+
  plt_pinpoint(0,0)

quad <- function(x,a,b,c){a*(x^2)+b*x+c} 

base + 
  geom_function(fun = quad,args = list(a = 1,b = 0, c = 0),
                linewidth = 1)+
  labs(title = expression(paste(y,' = ',x^2)))
  

```

In the example above $b$ and $c$ are zero, meaning that the function will display a simetric result around zero.

Now, let's say what happens when $a$ is negative.

```{r, echo = FALSE}
base + 
  geom_function(fun = quad,args = list(a = -1,b = 0, c = 0),
                linewidth = 1)+
  labs(title = expression(paste(y,' = -',x^2)))
```

In the example above $b$ and $c$ are zero, meaning that the function will display a simetric result around zero, but now the effect is inverted, where the values decreases as $x$ increases.

Now, let's say what happens when $b$ changes.

```{r, echo = FALSE}
base + 
  geom_function(fun = quad,args = list(a = 1,b = 0, c = 0),
                linewidth = 1, mapping = aes(colour = "a"))+
  geom_function(fun = quad,args = list(a = 1,b = 5, c = 0),
                linewidth = 1, mapping = aes(colour = "b"))+
  geom_function(fun = quad,args = list(a = 1,b = -5, c = 0),
                linewidth = 1, mapping = aes(colour = "d"))+
  scale_colour_manual(
    values = c('a' = 'black',
               'b' = 'royalblue3',
               "d" = "purple"),
    name = '',
    labels = expression(
      paste(y,' = ',x^2),
      paste(y,' = ',x^2,'+ 5x'),
      paste(y,' = ',x^2,'- 5x')
      )
    )

```

In the example above when $b > 0$, $y$ increases more for $x > 0$. At the same time when $b < 0$, $y$ increases more for $x < 0$.

Now, let's say what happens when $c$ changes.

```{r, echo = FALSE}
base+
  geom_function(fun = quad,args = list(a = 1,b = 0, c = 0),
                linewidth = 1, mapping = aes(colour = "a"))+
  geom_function(fun = quad,args = list(a = 1,b = 0, c = 50),
                linewidth = 1, mapping = aes(colour = "b"))+
  geom_function(fun = quad,args = list(a = 1,b = 0, c = -50),
                linewidth = 1, mapping = aes(colour = "d"))+
  scale_colour_manual(
    values = c('a' = 'black',
               'b' = 'firebrick3',
               "d" = "darkgoldenrod2"),
    name = '',
    labels = expression(
      paste(y,' = ',x^2),
      paste(y,' = ',x^2,'+ 50'),
      paste(y,' = ',x^2,'- 50')
      )
    )
```

In the example above we see that $c$ is just a incremental term, moving the function, but not changing its behavior.

# Proof

The goal of this post is to proof the formula of roots of the quadratic equation (i.e., Bhaskara formula), so:

$$
ax^2 + bx + c = 0.
$$ {#eq-quadratic-proof-01}

In the @eq-quadratic-proof-01, we divide by the term $a$:

$$
x^2 + \frac{bx}{a} + \frac{c}{a} = 0.
$$ {#eq-quadratic-proof-02}

In the @eq-quadratic-proof-02, we subtract the term $-\frac{c}{a}$:

$$
x^2 + \frac{bx}{a} = -\frac{c}{a}.
$$ {#eq-quadratic-proof-03}

Through a math property, we have that:

$$
(a+b)^2 = a^2+2ab + b^2.
$$ {#eq-math-01}

In the @eq-quadratic-proof-03, to achieve the property in @eq-math-01 we can add the term $\frac{b^2}{4a^2}-\frac{b^2}{4a^2}$:

$$
x^2 + \frac{bx}{a} +\left(\frac{b^2}{4a^2}-\frac{b^2}{4a^2}\right)= -\frac{c}{a}.
$$ {#eq-quadratic-proof-04}

In the @eq-quadratic-proof-04, we add the term $\frac{b^2}{4a^2}$:

$$
x^2 + \frac{bx}{a} +\frac{b^2}{4a^2}= -\frac{c}{a} + \frac{b^2}{4a^2}.
$$ {#eq-quadratic-proof-05}

In the @eq-quadratic-proof-05, we can see now that the left side is a case of @eq-math-01, then we apply it:

$$
\left(x + \frac{b}{2a}\right)^2= -\frac{c}{a} + \frac{b^2}{4a^2}.
$$ {#eq-quadratic-proof-06}

In the @eq-quadratic-proof-06, now we will put the right side in the same denominator

$$
\left(x + \frac{b}{2a}\right)^2= \frac{b^2-4ac}{4a^2}.
$$ {#eq-quadratic-proof-07}

In the @eq-quadratic-proof-07, we take the square root:

$$
x + \frac{b}{2a}= \pm \sqrt{\frac{b^2-4ac}{4a^2}}.
$$ {#eq-quadratic-proof-08}

In the @eq-quadratic-proof-08, we can apply the square root separately to the demonimator and numerator:

$$
x + \frac{b}{2a}= \pm \frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}.
$$ {#eq-quadratic-proof-09}

In the @eq-quadratic-proof-09, we can solve the denominator, since $\sqrt{4a^2} = 2a$.

$$
x + \frac{b}{2a}= \pm\frac{ \sqrt{b^2-4ac}}{2a}.
$$ {#eq-quadratic-proof-10}

In the @eq-quadratic-proof-10, we will move subtract the term $\frac{b}{2a}$.

$$
x = -\frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}.
$$ {#eq-quadratic-proof-11}

In the @eq-quadratic-proof-11, we just put all terms in the same denominator.

$$
x =  \frac{-b \pm\sqrt{b^2-4ac}}{2a}.
$$ {#eq-quadratic-proof-12}

Finally, we reached the Bhaskara formula in the @eq-quadratic-proof-12.