Prove why the Quadratic is what it is.

**JohnMon** · Mar 26th 2008, 02:59 PM

I need to learn why and how the Quadratic is what it is. Could their be anyway someone can show me the steps in it?

Thanks.

**topsquark** · Mar 26th 2008, 03:32 PM

Originally Posted by JohnMon

I need to learn why and how the Quadratic is what it is. Could their be anyway someone can show me the steps in it?

Thanks.

By "quadratic" do you mean you need to derive the quadratic formula?

This method uses "Completing the squares." If you are not comfortable with this method then I'd suggest you look at some examples first.

$ax^2 + bx + c = 0$

$ax^2 + bx = -c$

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

Now complete the square:
$a \left ( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2} \right ) = -c$

$a \left ( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} \right ) - a \cdot \frac{b^2}{4a^2} = -c$

$a \left ( x + \frac{b}{2a} \right ) ^2 = \frac{b^2}{4a} - c$

$a \left ( x + \frac{b}{2a} \right ) ^2 = \frac{b^2 - 4ac}{4a}$

$\left ( x + \frac{b}{2a} \right ) ^2 = \frac{b^2 - 4ac}{4a^2}$

$x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$

$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

There might be a "cooler" way to derive this, but I think this is the most "transparent" method. (The one that clearly shows where everything comes from.)

-Dan

**JohnMon** · Mar 26th 2008, 04:10 PM

Originally Posted by topsquark

By "quadratic" do you mean you need to derive the quadratic formula?

This method uses "Completing the squares." If you are not comfortable with this method then I'd suggest you look at some examples first.

$ax^2 + bx + c = 0$

$ax^2 + bx = -c$

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

Now complete the square:
$a \left ( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2} \right ) = -c$

$a \left ( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} \right ) - a \cdot \frac{b^2}{4a^2} = -c$

$a \left ( x + \frac{b}{2a} \right ) ^2 = \frac{b^2}{4a} - c$

$a \left ( x + \frac{b}{2a} \right ) ^2 = \frac{b^2 - 4ac}{4a}$

$\left ( x + \frac{b}{2a} \right ) ^2 = \frac{b^2 - 4ac}{4a^2}$

$x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$

$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

There might be a "cooler" way to derive this, but I think this is the most "transparent" method. (The one that clearly shows where everything comes from.)

-Dan

Thank you Dan!

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