# Problem with an algebra exercice

• Aug 14th 2010, 12:30 AM
pauline
Problem with an algebra exercice
Hello!!
I need your help with this exercise is about algebra i couldn't solve it
thank you so much
http://img41.imageshack.us/img41/7205/ninex.gif
Attachment 18558
and Attachment 18559
• Aug 14th 2010, 12:47 AM
DavidM
Expand and simplify the forumla and then use apply Quadratic formula to solve it.

$\displaystyle 3(x^2 - 4 \sqrt{3}) - 8x(\sqrt{3} - 1) + 19 = 0$
$\displaystyle 3x^2 - 12 \sqrt{3} - 8x\sqrt{3} - 8x + 19 = 0$
$\displaystyle 3x^2 - 8x\sqrt{3}-8x + 19 - 12\sqrt{3} = 0$

$\displaystyle a = 3$
$\displaystyle b = -8\sqrt{3} - 8$
$\displaystyle c = -12\sqrt{3} + 19$

$\displaystyle x = \dfrac{-b-\sqrt{b^2-4ac}}{2a}$

$\displaystyle x = \dfrac{-b+\sqrt{b^2-4ac}}{2a}$

(I may have got it wrong... it might have been moved to the calculus section for a reason)
• Aug 14th 2010, 09:20 AM
Tuufless
Completing the square
Are you familiar with how to complete the square? If so, then we rearrange your original equation,

$\displaystyle 3(x^2 - 4\sqrt{3}) - 8x(\sqrt{3} - 1) + 19 = 0$

into:
$\displaystyle 3x^2 - 8(\sqrt{3} - 1)x + 19 - 12\sqrt{3} = 0$

Factor out the 3:
$\displaystyle 3\left[x^2 - \frac{8(\sqrt{3} - 1)x}{3}\right] + 19 - 12\sqrt{3} = 0$

And complete the square:
$\displaystyle 3\left[\left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 - \left (\frac{4(\sqrt{3} - 1)}{3}\right )^2\right] + 19 - 12\sqrt{3} = 0$

Then some algebraic manipulation gets you (incoming wall-of-step-by-step-text):

$\displaystyle \left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 - \left (\frac{4(\sqrt{3} - 1)}{3}\right )^2 = \frac{12\sqrt{3} - 19}{3}$
$\displaystyle \left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 = \left (\frac{4(\sqrt{3} - 1)}{3}\right )^2 + \frac{12\sqrt{3} - 19}{3}$
$\displaystyle \left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 = \frac{16(4-2\sqrt{3})}{9} + \frac{36\sqrt{3} - 57}{9}$
$\displaystyle \left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 = \frac{64 - 32\sqrt{3} + 36\sqrt{3} - 57}{9}$
$\displaystyle \left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 = \frac{7 + 4\sqrt{3}}{9}$
$\displaystyle x-\frac{4(\sqrt{3}-1)}{3} = \pm\sqrt{\frac{7 + 4\sqrt{3}}{9}}$
$\displaystyle x-\frac{4(\sqrt{3}-1)}{3} = \pm\frac{\sqrt{7 + 4\sqrt{3}}}{3}}$

From here, notice that:
$\displaystyle 7+4\sqrt{3} = (2+\sqrt{3})^2$

Substituting that into the equation, we get:
$\displaystyle x-\frac{4(\sqrt{3}-1)}{3} = \pm\frac{\sqrt{(2+\sqrt{3})^2}}{3}}$

So,
$\displaystyle x=\frac{4(\sqrt{3}-1)}{3}\pm\frac{2+\sqrt{3}}{3}}$

Now, evaluate both solutions for x:
$\displaystyle x=\frac{4(\sqrt{3}-1)}{3}+\frac{2+\sqrt{3}}{3}}=\frac{5\sqrt{3}-2}{3}$
$\displaystyle x=\frac{4(\sqrt{3}-1)}{3}-\frac{2+\sqrt{3}}{3}}=\frac{3\sqrt{3}-6}{3}=\sqrt{3}-2$

Voila! Done!

If you're not sure how to complete the square, I'm sure there are other methods to doing this, such as using the quadratic equation as DavidM suggested. The "tricky" part is recognising that $\displaystyle 7+4\sqrt{3}$ can be expressed as a square in itself.
• Aug 14th 2010, 07:35 PM
pauline
thank you for your help both methods have helped me.