quadratic formula

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Sep 2nd 2009, 02:30 PM
mark

quadratic formula

i'm trying to use the quadratic formula for $3x^2 - 12x + 5 = 0$

and got it to $x = \frac {12 \pm \sqrt {-12^2 - 60}}{6}$ not sure if i'm going about it right so would like someone to point me in the right direction if possible

thanks
Sep 2nd 2009, 02:39 PM
I-Think

Yes you are very correct.
But my teacher insisted on being picky so in that sentiment I'll just say that your notation, instead of http://www.mathhelpforum.com/math-he...dac4037a-1.gif

Should be

$x=\frac{12\pm{\sqrt{(-12)^2-60}}}{6}$

The addition of brackets removes confusion, that's what my teacher used to say.
Sep 2nd 2009, 02:44 PM
mark

ah thanks, the answer the book gives is $x = 2 \pm \sqrt {\frac{7}{3}}$ so i can't see how you'd arrive at that from the answer i was at, could you show me the way to get to it please? there are also meant to be rackets under the root sign around the 7/3 like so: (7/3) but i don't know how to write it out using the brackets like that
Sep 2nd 2009, 02:57 PM
skeeter

Quote:

Originally Posted by mark

i'm trying to use the quadratic formula for $3x^2 - 12x + 5 = 0$

and got it to $x = \frac {12 \pm \sqrt {-12^2 - 60}}{6}$ not sure if i'm going about it right so would like someone to point me in the right direction if possible

thanks

$a = 3$

$b = -12$

$c = 5$

$x = \frac{12 \pm \sqrt{(-12)^2 - 4(3)(5)}}{2(3)}$

$x = \frac{12 \pm \sqrt{144 - 60}}{6}$

$x = \frac{12 \pm \sqrt{84}}{6}$

$x = \frac{12 \pm \sqrt{4 \cdot 21}}{6}$

$x = \frac{12 \pm 2\sqrt{21}}{6}$

$x = \frac{6 \pm \sqrt{21}}{3} = 2 \pm \frac{\sqrt{21}}{3}$

note that $\sqrt{\frac{7}{3}} = \frac{\sqrt{7}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{21}}{3}$