Rearranging the quadratic formula

Hi guys, I've been tasked with rearranging the quadratic root formula to make "a" the subject. I've been struggling on this one for a little while now and can't figure out how to isolate a. For reference, the equation i'm referring to is the addition half of the formula, i.e.

x = (-b + sqroot(b^2 - 4ac))/2a

Any help is much appreciated.

Regards,

Sathya

Re: Rearranging the quadratic formula

Because that formula is a root of the equation we know that ax^2+bx+c=0 So ax^2=-bx-c So a=(-bx-c)/x^2

Re: Rearranging the quadratic formula

Thanks very much, I thought it work be a little more complex than that to be honest, but thanks for showing me :)

Very speedy response, cheers.

Re: Rearranging the quadratic formula

Hello, SSmith!

biffboy has an excellent solution.

I'll "do" the algebra.

Quote:

I've been tasked with rearranging the quadratic root formula to make $\displaystyle a$ the subject.

. . $\displaystyle x \:=\: \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\quad a \ne 0$

We have: .$\displaystyle x \:=\:\frac{-b \pm\sqrt{b^2-3ac}}{2a}$

Mulitply by $\displaystyle 2a\!:\;\;2ax \;=\;-b \pm\sqrt{b^2-4ac} $

Add $\displaystyle b\!:\;\;2ax + b \;=\;\sqrt{b^2-4ac}$

Square: .$\displaystyle (2ax + b)^2 \;=\;\left(\sqrt{b^2-4ac}\right)^2$

. . $\displaystyle 4a^2x^2 + 4abx + b^2 \;=\;b^2 - 4ac$

. $\displaystyle 4a^2x^2 + 4abx + 4ac \;=\;0$

. . .$\displaystyle 4a(ax^2 + bx + c) \;=\;0$

And we have two equations to solve.

$\displaystyle 4a \:=\:0 \quad\Rightarrow\quad a \:=\:0\quad\text{ not allowed!}$

$\displaystyle \underbrace{ax^2 + bx + c \:=\:0} \quad\Rightarrow\quad ax^2 \:=\:-bx - c \quad\Rightarrow\quad a \:=\:\frac{-bx-c}{x^2}$

biffboy started here!

Re: Rearranging the quadratic formula

Thank you so much for that extra clarification, that really helps me to understand how to get back to standard form, much appreciated Soroban!

Re: Rearranging the quadratic formula

this is the only confirmation that i can find on the web about subject "a" in the quadratic formula! thanks!