Finding Minimum from a completion of a square...

[laker_gurl3]

so the question is:

Complete the square of x^2 + 4x - 1 and hence find the position and value of its minimum.

AFter completing the square i have : (x+2)^2 - 5 ,

How do I find the minimum value from that?

Thanks in advance guys!

[Tide]

What is the smallest possible value for (x+2)^2?

[laker_gurl3]

umm so would x= -2 and y = -5

??

[TD]

Indeed :smile:

[laker_gurl3]

okaay another questoin....
Factorise and hence solve the equation:

2x^3 + 3x^2 - 8x - 12 = 0

I can't seem to remember how to factorize this because there's a cubic expression! how do i start?

[TD]

Divisors of the constant term (i.e. -12 here) are possible zeroes of your equation. Try to find such a zero and use the fact that if x = a is a zero, you can factor out (x-a).

[laker_gurl3]

okay thanks, i'm going to try that now, by the way is that what you call the remainder theorem?

[laker_gurl3]

Okay, so my answer was:

(x-2)(x+2)(2x+3)

is that right?

[Tide]

Okay, so my answer was:
(x-2)(x+2)(2x+3)
is that right?

You can expand that product and see if it produces your original expression.

[laker_gurl3]

So there's this question to factorize:

x^4 - 3x^2 - 10 = 0

How do I start this one because I can't find a factor to go into it.

[TD]

Okay, so my answer was:
(x-2)(x+2)(2x+3)
is that right?
Yes, which is easy to see if you follow Tide's advice.

So there's this question to factorize:
x^4 - 3x^2 - 10 = 0
How do I start this one because I can't find a factor to go into it.
Would it be easier for you to see if you let x^2 = t so it becomes quadratic?

That would give t^2-3t-10=0 to factor and then substitute t by x² again.

[laker_gurl3]

so the ans is \sqrt{5} ?

[Tide]

so the ans is \sqrt{5} ?

That's one of them! :)

[laker_gurl3]

Ohh is it +/- Square root 5?! lol

[TD]

Ohh is it +/- Square root 5?! lol
Indeed, unless you're working complex since then a quartic equation has 4 solutions. The ones you found are the only real ones though.

[laker_gurl3]

help with this last one

Express the function \sqrt{3}\sin{2t} - 3\cos{2t} in the form A\sin{(2t+\alpha)


i have no idea what to do

[Tide]

Ohh is it +/- Square root 5?! lol

Now you're just guessing!

t^2 - 3t - 10 = (t-5)(t+2)

[TD]

Yes, but t was x².

[laker_gurl3]

well if you sub the x^2 back in to the equation i get \pm\sqrt{5}. is that not correct? and what of the other question? how should i start what am i suppose to do?

[TD]

I would 'expand' sin(2t+a) with the formula for addition of angles, being:

\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b

[laker_gurl3]

ya i tried to do that but i did not equal the left side.... so....? am i doin something wrong?

[TD]

It won't be equal to it immediately but some parts will be the same. Identify the other factors with the ones which were there in the given expression.

[laker_gurl3]

ok im completely lost now heres what i got:

\sin{2t}\cos{\alpha} + \cos{2t}\sin{\alpha}

theres nothing more that i can do to make it look like the right side.

[TD]

Don't forget the factor a that was in front of that sine at the RHS.
Then compare the two terms with those on the LHS. What is already equal and which parts do you have to 'make' equal?

[laker_gurl3]

??how do i factor A ?i dont see anything that is equal on the RHS...

[TD]

So at the RHS, using that formula gives:

a\sin \left( {2t + \alpha } \right) = a\sin 2t\cos \alpha + a\cos 2t\sin \alpha

Let's compare that to the LHS:

\sqrt 3 \sin 2t - 3\cos 2t = a\sin 2t\cos \alpha + a\cos 2t\sin \alpha

In the first term, we have a sin(2t) at both sides. In the second term, we have a cos(2t) at both sides. If we can get the other parts to equal to, we're there.

For the first term, this means that a\cos \alpha = \sqrt 3.
Do you see that? Can you check what that gives for the second term?

[laker_gurl3]

wait how did u get a\cos{\alpha} = \sqrt{3} ??

[TD]

I'm only writing the first term of both sides now:

\boxed{\sqrt 3 }\sin 2t = \boxed{a\cos \alpha }\sin 2t

As you can see, the sin(2t) is already there on both sides so of course, that's equal. In order for the entire thing to be equal, the other factors have to be equal as well. Do you see that this gives the equation you just mentioned?

[laker_gurl3]

ok so now that we have a\cos{\alpha}\sin{2t} = \sqrt{3}\sin{2t} what happens then? what am i suppose to do with the \sqrt{3} = a\cos{\alpha}

am i suppose to isolate the a???

[TD]

Find another equation by comparing the second terms of each side.
Then you have 2 equations and 2 unknowns (a and alpha).

[laker_gurl3]

\sqrt 3 \sin 2t = a\sin 2t\cos \alpha

is that the second equation?

the asin 2t and the sqrt 3 ?

[TD]

Comparing the second terms would look like this:

\boxed{ - 3}\cos 2t = \boxed{a\sin \alpha }\cos 2t

[laker_gurl3]

what happened to the \sqrt{3} ? and how come its like that now?

[TD]

Let's try again, we had:

\underbrace {\sqrt 3 }_A\underbrace {\sin 2t}_B\underbrace { - 3}_C\underbrace {\cos 2t}_D = \underbrace {a\cos \alpha }_{A'}\underbrace {\sin 2t}_{B'} + \underbrace {a\sin \alpha }_{C'}\underbrace {\cos 2t}_{D'}

I named all the parts, A,B,C,D at the LHS and the same parts in the RHS with a '. It is clear that B = B' and that D = D'. Now if we let A = A' and C = C', we have what we want. Do you see that?

That would give the system of the following two equations:


\left\{ \begin{gathered}
a\cos \alpha = \sqrt 3 \hfill \\
a\sin \alpha = - 3 \hfill \\
\end{gathered} \right.

[laker_gurl3]

yes i see that but what are we going to do with

\sqrt{3} = a\cos{\alpha} and -3 = a\sin{\alpha}


what can you do with it?

[TD]

Diving both equations would give \tan \alpha = - \sqrt 3 from which you can get alpha.

[laker_gurl3]

but if u divide them both it will be

a\tan{\alpha} = -\frac{\sqrt{3}}{3}

[TD]

The a will cancel out and what you wrote isn't correct. You either divide the first by the second and get \cot \alpha = - \sqrt 3 /3 or you divide the second by the first and you get what I had, which has the same solutions for alpha of course :smile:

[laker_gurl3]

so what is the final answer? \tan \alpha = - \sqrt 3
and thats it?

[TD]

Now find alpha...

[laker_gurl3]

what do you mean find alpha? didnt we just do that? is that -60 degrees or \frac{-\pi}{3}

its just that i have no idea on what the final equation looks like both sides dont match at all right now, i have:

\sqrt{3}\sin{2t} - 3\cos{2t} = -\frac{\pi}{3}

is that correct so far?

[TD]

Yes, or 2pi/3. Now it's easy to find a using one of the two equations.

[laker_gurl3]

wait you said you can find A by using one of the two equations, you mean like this

a\sin{2pi/3} = -3


???

[TD]

Indeed, now solve for a.

[laker_gurl3]

i would get -3.464 now what do i do with that? can you just tell me the answer its been 7 hours already...

[TD]

sin(2pi/3) = sqrt3/2, so a = -2sqrt3 when you use that angle alpha.
You can find another solution by using the other angle.

[laker_gurl3]

what other angle?!!! we only have -60

[TD]

\tan \alpha = - \sqrt 3 \Leftrightarrow \alpha = - \pi /3 + 2k\pi \,\, \vee \,\, \alpha = 2\pi /3 + 2k\pi \Leftrightarrow \alpha = 2\pi /3 + k\pi

[laker_gurl3]

.....this is awsome almost 8 hours and still no answer

[TD]

Perhaps you should re-read all of the posts and think carefully, all the necessary information has been said. If something is still unclear, be specific about what the problem actually is.

[laker_gurl3]

after all this 50 some post....look at the progression...

\sqrt{3}\sin{2t} - 3\cos{2t} = -\frac{\pi}{3}

seems to me that this is gettin further away from the answer...

[TD]

Seems to me that you have not carefully read everything which has been said. The -pi/3 was a solution for alpha, not for the RHS in general.