1. Im stuck!

'm stuck! I'm trying to find x using the cosine rule from:

The answer is; x = 7.23 from:

How?

In another question i have to find the value of x which has the minimum value:

$\displaystyle b^2 = x^2 + x + 7$

I'm guessing i have to complete the square? but $\displaystyle b = b^2$ so do i take the route of the whole thing?

2. For the first question, use the quadratic formula. The solutions for $\displaystyle ax^2 + bx + c = 0$ are $\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

For the second question, completing the square is a good idea. You will get $\displaystyle b^2 = \left(x + \frac{1}{2}\right)^2 + \frac{27}{4}$.

Then the minimum value for b is $\displaystyle \sqrt{\frac{27}{4}}$.

3. i don't understand how to get the $\displaystyle x^2 - 4(\sqrt2-1)x - (29+8\sqrt2)$

And the second one gives the answer to be $\displaystyle {\frac {1}{2}}$?

4. Originally Posted by Flexible
i don't understand how to get the $\displaystyle x^2 - 4(\sqrt2-1)x - (29+8\sqrt2)$

And the second one gives the answer to be $\displaystyle {\frac {1}{2}}$?
The solutions for $\displaystyle x^2 - 4(\sqrt2-1)x - (29+8\sqrt2) = 0$ are:

$\displaystyle x = \frac{4(\sqrt2 - 1) \pm \sqrt{(4(\sqrt2 - 1))^2 + 4(1)(29+8\sqrt2)}}{2}$.

For the second problem, the answer is $\displaystyle x = -\frac{1}{2}$.

5. How do you get $\displaystyle -\frac{1}{2}$ I.e $\displaystyle \sqrt\frac{27}{4} = -\frac{1}{2}$ How?