• Jan 17th 2011, 07:36 AM
andyboy179
hi, i have a quadratic equation and i have done it but im not sure its correct so if someone could check i will be very grateful, heres the question:

Attachment 20475

this is what i got:
Attachment 20476

is this correct? if not where am i going wrong.
• Jan 17th 2011, 07:41 AM
harish21
correct
• Jan 17th 2011, 07:42 AM
Prove It

$\displaystyle 3x^2 - 4x + 1 = 0$

$\displaystyle 3x^2 - 3x - x + 1 = 0$

$\displaystyle 3x(x - 1) - 1(x - 1)= 0$

$\displaystyle (x - 1)(3x - 1) = 0$

$\displaystyle x -1 = 0$ or $\displaystyle 3x - 1 = 0$

$\displaystyle x = 1$ or $\displaystyle x = \frac{1}{3}$.
• Jan 17th 2011, 11:30 AM
ramdrop
Quote:

Originally Posted by harish21
correct

Yes it is, but here is an easier way without the formula
You can do this:

$3x^2 -4x + 1 = 0$

So if you know how to do quadratic equations, you must have a 3 and a 1 in the brackets to achieve the $3x^2$

$(3x ...)(x ...)$
Obviously the factors of 1, is 1 and itself so you can get:
$(3x .. 1)(x .. 1)$
Now about the signs, you need a + 1, so you need either 2 +'s or 2 -'s, seeing as you have -4 in the middle you can achieve this with 2 -'s

so:
$(3x - 1)(x - 1) = 0$
$x = 1 or x = 1/3$

Regards
• Jan 17th 2011, 01:18 PM
HallsofIvy
Strictly speaking, "0.33" is NOT a correct answer- $\frac{1}{3}$ is.
0.33 is only approximately equal to 1/3.

Also, you should understand that you could have checked these by simply substituting you answers into the given equation.

If x= 1, then $3x^2- 4x+ 1= 3(1)^2- 4(1)+ 1= 3- 4+ 1= 0$ and $3\left(\frac{1}{3}\right)^2- 4\left(\frac{1}{3}\right)+ 1= \fra{1}{3}- \frac{4}{3}+ 1= 0$.

And, as others have pointed out, if a polygon can be factored, that is a better way of solving it than using a formula.
• Jan 17th 2011, 01:42 PM
e^(i*pi)
Quote:

Originally Posted by HallsofIvy
And, as others have pointed out, if a polygon can be factored, that is a better way of solving it than using a formula.

If $a \neq 1$ I usually bang out the formula to save time. I don't really think there is a preferred way (unless the question demands it of course)