Problem with an algebra exercice

**pauline** · August 14th 2010, 12:30 AM

Hello!!
I need your help with this exercise is about algebra i couldn't solve it
thank you so much

the answers are:

Problem with an algebra exercice-nine2.gif

and

Problem with an algebra exercice-nine3.gif

**DavidM** · August 14th 2010, 12:47 AM

Expand and simplify the forumla and then use apply Quadratic formula to solve it.

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

Quadratic formula:

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

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

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

Substitute in the values, simplify, and you get your answer.

(I may have got it wrong... it might have been moved to the calculus section for a reason)

**Tuufless** · August 14th 2010, 09:20 AM

Are you familiar with how to complete the square? If so, then we rearrange your original equation,

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

into:

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

Factor out the 3:

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

And complete the square:

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

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

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

From here, notice that:

$7+4\sqrt{3} = (2+\sqrt{3})^2$

Substituting that into the equation, we get:

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

So,

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

Now, evaluate both solutions for x:

$x=\frac{4(\sqrt{3}-1)}{3}+\frac{2+\sqrt{3}}{3}}=\frac{5\sqrt{3}-2}{3}$
$x=\frac{4(\sqrt{3}-1)}{3}-\frac{2+\sqrt{3}}{3}}=\frac{3\sqrt{3}-6}{3}=\sqrt{3}-2$

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 $7+4\sqrt{3}$ can be expressed as a square in itself.

**pauline** · August 14th 2010, 07:35 PM

thank you for your help both methods have helped me.

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