I am attempting to derive the quadratic formula by completing the square on the generic generic expression:
$$ax^2+bc+c=0$$
I'm struggling with the right hand side of the equation which, for the step I'm on I know should be $\frac{b^2-4ac}{4a^2}$. However, I arrive at $\frac{b^2a-4a^2c}{4a^3}$
Here's my working:
(Approach copied largely from textbook)
Start with:
$ax^2+bx+c=0$
Move constant term to the right:
$ax^2+bx=-c$
Divide by $a$ to ensure leading coefficient is 1:
$x^2+\frac{b}{a}x=-\frac{c}{a}$
Calculate the amount needed to complete the square and add to both sides:
$(\frac{1}{2}\times \frac{b}{a})^2$ = $(\frac{b}{2a})^2$ = $\frac{b^2}{4a^2}$
Now add this to both sides:
$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}+-\frac{c}{a}$
Write the left side as a perfect square:
$(x^2+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{c}{a}$
Simplify the right hand side by finding a common denominator:
This is where I'm tripping up
$\frac{b^2}{4a^2}-\frac{c}{a}$
The common denominator will be the product of the denominators so $4a^3$
This doesn't "feel" right and I suspect I should be looking for a "least common denominator" but I don't know what that would be given the existence of the radical.
Rewriting using the common denominator $4a^3$ I multiply the numerator and denominator of left side of the minus sign by just $a$. I then multiple the numerator and denominator on the right side of the minus sign by $4a^2$:
$\frac{b^2a}{4a^3}-\frac{4a^2c}{4a^3}$ = $\frac{b^2a-4a^2c}{4a^3}$
How can I arrive at $\frac{b^2-4ac}{4a^2}$?
I know that I'm not done yet after figuring out the above, but it's this in between step I'm tripping up on.
Best Answer
"The common denominator will be the product of the denominators so $4a^3$" -- this is wrong.
The common denominator is $4a^2$, not $4a^3$. You multiply denominators only when they don't have a common factor; here they do.