Rearranging the quadratic formula

**SSmith** · April 14th 2012, 02:26 AM

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

**biffboy** · April 14th 2012, 02:55 AM

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

**SSmith** · April 14th 2012, 05:51 AM

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.

**Soroban** · April 14th 2012, 06:11 AM

Hello, SSmith!

biffboy has an excellent solution.
I'll "do" the algebra.

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

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

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

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

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

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

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

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

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

And we have two equations to solve.

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

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

**SSmith** · April 14th 2012, 04:49 PM

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

**phillphall** · December 10th 2012, 03:38 PM

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

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