What do the constants "4" and "2" in Bhaskara mean and where did they come from?

Question

In bhaskar, the way to get the result, is to get the $\Delta = b^2 – 4ac$, and then the $X = (–b \pm \sqrt\Delta)/2a)$. But from where come these constants?

Answer 1

Begin with

$$ax^2 + bx + c = 0.$$

Divide out $a$:

$$x^2 + \frac{bx}{a} + \frac{c}{a} = 0.$$

Complete the square:

$$x^2 + \frac{bx}{a} + \frac{b^2}{(2a)^2} =-\frac{c}{a} + \frac{b^2}{(2a)^2}.$$

Can you take it from here?

Answer 2

The goal is to get $ax^2 + bx + c = 0$ into a form of $(x+k)^2 = m$ and thus we can solve:

$x+k =\pm m$

$x = -k\pm m$.

and we do that by "completing the square".

First to go from $ax^2 + bx + c=0$ to $(x+k)^2 =m$ we need to get rid of the leading coefficient, $a$....

So we divide both sides by $a$ (presumably $a\ne 0$ [otherwise it's not a quadratic equation]) to get:

$x^2 + \frac bax + \frac ca =0$.

Let to get this into the form $(x+k)^2 = m$ we note that if we expand $(x+k)^2=m$ we get that $(x+k)^2 = (x+k)(x+k)$ and First Outside Inside Last we have $(x+k)(x+k) = x^2 + kx + kx + k^2= x^2 + 2kx + k^2$.

Two things to note (since yous asked). In multiplying the insides and the outsides, the inside and the outsides are the same si we get the term $\color{red}2kx$. ANd that's the first appearence of the $\color{red}2$.

Now we want to get from $x^2 + \frac ba x + c =0 \iff x^2 + 2kx + k^2 = m$. and that requires the $x^2$ terms be the same, that the $x$ terms be the same, and that when we combine the constant terms to one side of the equation or the other, that they be the same.

The $x^2$ are already the same. But the $x$ terms are in our equation is $\frac ba$ and we want it to be $2k$ so we must have $\frac ba = 2k$ and we must have $k = \frac b{2a}$ so we can put the $x$ term $x^2 +\frac ba x + c =0$ into that $x^2 +2kx +k^2 = m$ form. We have

$x^2 + \color{red}2\frac {b}{\color{red}2a} + c = 0$.

That is where the $2$ comes from.

Now we need to come up with the $k^2$ constant term. As $k =\frac b{2a}$ then $k^2 = (\frac b{2a})^2 = \frac {b^2}{\color{blue}4ra^2}$ and that is what the $\color{blue}4$ comes from.

Now to get $x^2 + \color{red}2\frac {b}{\color{red}2a} + c = 0$ to $x^2 + 2kx + k^2 = m$ we need to replace the "$.....+ c = 0$ with "$..... + k^2 = m$.

First get the $c$ out of the way be subtracting $c$ from both sides

$x^2 + \color{red}2\frac {b}{\color{red}2a} + c = 0$

$x^2 + \color{red}2\frac {b}{\color{red}2a} = -c$.

But now we need to introduce the $k^2$ part in. $k^2 = \frac {b^2}{\color{blue}4a^2}$ so we add $\frac {a^2}{\color{blue}4b^2}$ to both sides.

$x^2 + \color{red}2\frac {b}{\color{red}2a}+\frac {b^2}{\color{blue}4a^2} = -c+\frac {b^2}{\color{blue}4a^2}$

And for neatness we put the RHS over a common denominator:

$x^2 + \color{red}2\frac {b}{\color{red}2a}+\frac {b^2}{\color{blue}4a^2} = -c+\frac {b^2}{\color{blue}4a^2}=\frac {-c\color{blue}4a^2 + b^2}{\color{blue}4a^2}=\frac {b^2-\color{blue}4a^2c}{\color{blue}4a^2}$

Now we (finally) have it in $x^2 + 2k + k^2 = m$ form. So to put it in $(x+k)^2 = m$ we consolidate the square.

$x^2 + \color{red}2\frac {b}{\color{red}2a}+\frac {b^2}{\color{blue}4a^2} =\frac {b^2-\color{blue}4a^2c}{\color{blue}4a^2}$

$(x + \frac {b}{\color{red}2a})^2 = \frac {b^2-\color{blue}4a^2c}{\color{blue}4a^2}$

Now solve for $x$.

$x + \frac {b}{\color{red}2a}=\pm \sqrt{\frac {b^2-\color{blue}4a^2c}{\color{blue}4a^2}}=\pm \frac{\sqrt{b^2-\color{blue}4a^2c}}{\color{red}2a}$

$x =-\frac {b}{\color{red}2a}\pm \frac{\sqrt{b^2-\color{blue}4a^2c}}{\color{red}2a}$

$x = \frac{-b \pm\sqrt{b^2-\color{blue}4a^2c}}{\color{red}2a}$