Im stuck!

**Flexible** · October 9th 2008, 10:23 AM

'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:

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

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

**icemanfan** · October 9th 2008, 10:27 AM

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

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

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

**Flexible** · October 9th 2008, 10:34 AM

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

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

**icemanfan** · October 9th 2008, 10:59 AM

Originally Posted by Flexible

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

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

The solutions for $x^2 - 4(\sqrt2-1)x - (29+8\sqrt2) = 0$ are:

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

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

**Flexible** · October 9th 2008, 11:13 AM

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

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