quadratic functions

[Guest]

Hi, I hope I am writing these problems correctly:

8/ sqrt 60= ?

-3(4x-9)^2+5=-3
The book said the answer is 9+ or - 2 sqrt 3/ 4 (Again, sorry if this is not written in correct format.)

 

[tkhunny]

Hi, I hope I am writing these problems correctly:

8/ sqrt 60= ?

-3(4x-9)^2+5=-3
The book said the answer is 9+ or - 2 sqrt 3/ 4 (Again, sorry if this is not written in correct format.)

You are not writing them correctly because you have included no problem statement. What is to be done?

Let's assume the first is "Simplify."

$\displaystyle \frac{8}{\sqrt{60}} = \frac{8}{\sqrt{4*15}} = \frac{8}{2*\sqrt{15}} = \frac{4}{\sqrt{15}} = \frac{4}{\sqrt{15}}*\frac{\sqrt{15}}{\sqrt{15}} = \frac{4}{15}*\sqrt{15}$

That's about it.

 

[Unco]

For the first one, \(\displaystyle \frac{8}{\sqrt{60}\), I assume you want to simplify it?

Well, $\displaystyle \sqrt{60} = \sqrt{4\cdot15} = \sqrt{4}\cdot\sqrt{15} = 2\sqrt{15}$, so you end up with $\displaystyle \frac{4}{\sqrt{15}}$

Note that the radical, $\displaystyle \sqrt{blah}$ denotes to the positive square root (of "blah").

For the second one, you have written the question fine (well, it's unambiguous, anyway).

-3(4x - 9)^2 + 5 = -3

Move that 5 to the right-hand side by subtracting both sides by 5:

-3(4x - 9)^2 = -8

Divide both sides by -3:

$\displaystyle (4x - 9)^2 = \frac{8}{3}$

Take square roots of both sides:

$\displaystyle 4x - 9 = \pm \sqrt{\frac{8}{3}}$

Move the -9 to the right-hand side:

$\displaystyle 4x = 9 \pm \sqrt{\frac{8}{3}}$

Divide both sides by 4:

$\displaystyle x = \frac{9 \pm \sqrt{\frac{8}{3}}}{4}$

We can pay around with the radicals:

$\displaystyle \sqrt{\frac{8}{3}} = \frac{\sqrt{8}}{\sqrt{3}}$

$\displaystyle = \frac{\sqrt{4\cdot2}}{\sqrt{3}} = \frac{2\sqrt{2}}{\sqrt{3}}$

And you end up with:

$\displaystyle x = \frac{9}{4} \pm \frac{\sqrt{2}}{2\sqrt{3}}$

Your version of the provided answer, I think, translates to:

$\displaystyle \frac{9 \pm 2\sqrt{3}}{4}$

... so perhaps something was lost in translation?

Edit: Looking at my answer, I think it would be best to rationlise the denominator:

$\displaystyle \frac{\sqrt{2}}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{6}}{6}$