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Commentary for "Method for Checking the Solutions to a Quadratic"

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
temp

Moderator edit: This commentary topic pertains to the following tutorial:

http://mathhelpboards.com/math-notes-49/method-checking-solutions-quadratic-4101.html
And for anyone wondering how we get the Quadratic Formula, it's just completing the square on a standard quadratic equation \(\displaystyle \displaystyle a\,x^2 + b\,x + c = 0\).

\(\displaystyle \displaystyle \begin{align*} a\,x^2 + b\,x + c &= 0 \\ a \left( x^2 + \frac{b}{a}\,x + \frac{c}{a} \right) &= 0 \\ x^2 + \frac{b}{a}\,x + \frac{c}{a} &= 0 \\ x^2 + \frac{b}{a}\,x + \left( \frac{b}{2a} \right) ^2 - \left( \frac{b}{2a} \right) ^2 + \frac{c}{a} &= 0 \\ \left( x + \frac{b}{2a} \right) ^2 - \frac{b^2}{4a^2} + \frac{4ac}{4a^2} &= 0 \\ \left( x + \frac{b}{2a} \right) ^2 &= \frac{b^2 - 4ac}{4a^2} \\ x + \frac{b}{2a} &= \frac{\pm \sqrt{b^2 - 4ac}}{2a} \\ x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{align*}\)
 
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Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
Re: Method for Checking the Solutions to a Quadratic

And for anyone wondering how we get the Quadratic Formula, it's just completing the square on a standard quadratic equation \(\displaystyle \displaystyle a\,x^2 + b\,x + c = 0\).

\(\displaystyle \displaystyle \begin{align*} a\,x^2 + b\,x + c &= 0 \\ a \left( x^2 + \frac{b}{a}\,x + \frac{c}{a} \right) &= 0 \\ x^2 + \frac{b}{a}\,x + \frac{c}{a} &= 0 \\ x^2 + \frac{b}{a}\,x + \left( \frac{b}{2a} \right) ^2 - \left( \frac{b}{2a} \right) ^2 + \frac{c}{a} &= 0 \\ \left( x + \frac{b}{2a} \right) ^2 - \frac{b^2}{4a^2} + \frac{4ac}{4a^2} &= 0 \\ \left( x + \frac{b}{2a} \right) ^2 &= \frac{b^2 - 4ac}{4a^2} \\ x + \frac{b}{2a} &= \frac{\pm \sqrt{b^2 - 4ac}}{2a} \\ x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{align*}\)
Right, or you can do a couple other equivalent methods.