Understanding Algebra 1: Multiplying (AX + B)(C - D) Simplified

[pointintime]

the answer is 3.111

don't know how to get it

 

[belliott4488]

-2^-1 t3^2 - 180 s (t3) + t3^2 = a^-1 (Vo (180 s) - 1100 m + Xo)
(Vo (180 s) - 1100 m + Xo)^-1 a(-2^-1 t3^2 - 180 s (t3) + t3^2) = 0

:O

-2^-1 t3^2 - 180 s (t3) + t3^2

ok how do I combine

-2^-1 t3^2

and

t3^2

:O

Okay, don't take this the wrong way, but you really need to brush up on some basic algebra.

Let's write out your equation a little more clearly. We start with:

-2^-1 t3^2 - 180 s (t3) + t3^2 = a^-1 (Vo (180 s) - 1100 m + Xo)

Now first of all, do you know that any number to the power -1 is just the reciprocal of that number? There's no reason to write "-2^-1" -- that's just -1/2. Same with a^-1 -- that's just 1/a. So let's start with:

-1/2 t3^2 - 180 s (t3) + t3^2 = 1/a (Vo (180 s) - 1100 m + Xo)

Now, you need to identify the terms, that is, the quantities that are being added or subtracted. Those are the ones that you can "move" to the other side of the equal sign by adding or subtracting them from both sides. You have three terms on the left side and one term on the right side. We'll get them all on the left side by subtracting the right side from both sides of the equation. That gives us:

-1/2 t3^2 - 180 s (t3) + t3^2 - 1/a (Vo (180 s) - 1100 m + Xo) = 0

Follow that?

Now, you need to identify the terms with t3^2, with t3, and with no t3 at all. There are four terms on the left side now, but we should not have more than three, so you must find two that have the same power of t3 and add them.

You identified the two terms that have t3^2 in the second quote above. Now that I've rewritten them as -1/2 t3 and t3^2, can you combine them? You have one t3^2 and you're adding a negative 1/2 t3^2, which is the same as subtracting 1/2 t3^2.

 

[pointintime]

ok I followed that let me work it

 

[pointintime]

so this is it?

1/2 t3^2 - 180 s (t3) - 1/a (Vo (180 s) - 1100 m + Xo) = 0

 

[pointintime]

so now I do the opposte of b plus or minus...

sense a is 1/2 it becomes 2 correct? I think i got it

 

[pointintime]

please tell me this is correct

t3 = (.20 m/s(180 s +/- sqrt((-180 s)^2 - (4)((5.494 m/s )(180 s) -1,100 m))/(2(-.20 m/s^2)))/2

 

[belliott4488]

so now I do the opposte of b plus or minus...

sense a is 1/2 it becomes 2 correct? I think i got it

Yeah, you're on your way now - just be careful because you've got two a's you're dealing with. There's the a in the original equation and the a in the quadratic formula - they're not the same. It might help to use A, B, B for the general form of the quadratic equation and the quadratic formula, just so you don't get confused.

The denominator is 2A, and yes, A = 1/2, but what is 2 * 1/2? It's not 2 ...

 

[pointintime]

It's 1...

ok I calculate C to be 555.4 m/s^2
C = (5.494 m/s (180 s) -1,110 m)/-(.20 m/s^2) when I put that into my calculator I got
555.4 not sure on units here

180 +/- ( (-180 s)^2 - (4(555.4 m/s^2))/2)^(.5)

and I got ridiculous answers


Also my units didn't cancel out =[

I got

356.88753 I know it's seconds but can't prove units don't cancel out

and

3.112465 again know it's seconds but can't prove

 

[HallsofIvy]

ok yes just got anohter question...

Ok well I got down to this and don't know what to do please someone help me...

-2^-1 t3^2 - 180 s t3 + t3^2 = a^-1 (Vo (180 s) - 1,100 m + Xo)

$$-2^{-1}t_3^2- 180 t_3+ t_3^2= a^{-1}(180V_0- 1100+ X_0$$
is the same as
$$(-\frac{1}{2}+1)t_3^2- 180t_3- \frac{180V_0- 1100+ X_0}{a}= 0$$
$$\frac{1}{2}t_3^2- 180t_3- \frac{180V_0- 1100+ X_0}{a}= 0$$
You can solve that using the quadratic formula. Of course, you will have $V_0$, $X_0$, and a inside the square root. If you want a numerical answer you will have to supply values for them.

how do I solve this for t3

the 3 is a subscript to t
180 s is 180 seconds and is considered one term
Vo is one term
Xo is one term

 

[pointintime]

I put my answers in 43

Vo = 5.494 m/s

a = .20 m/s^2

 

[pointintime]

I think I have solved this problem

THANK YOU!