The trick here is to factor it and simplify it before you do any multiplication.

The length is given right over there it is a squared plus 6 a minus 27 they want us to simplify this.

So I moved to the Geneva University.

And that can also be written as and this is my preferred notation, because it doesn't have all this extra junk everywhere.

So what does this simplify to?

And all of these are divisible by 5.

So let's multiply both sides by mu.

So if you have the opportunity, please try it out.

Times v1 dot v1.

So this whole expression simplifies to

Is given right up here, is 2x times the principal square root of 5 minus 4, minus 4.

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And then you make some assumptions about it.

So let's factor out a 6. And then 12x divided by 6 is 2x. And then 30 divided by 6 is 5.

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It's round, it's neat.

We've seen how expensive it would be to track all the photons in a scene.

So that's our operationalization, of verbal ability.

We have one expression on top of another expression.

The numerator has the same degree as the denominator.

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I'm starting to see the light at the end of the tunnel.

let's cancel that out.

25i 2 is the same thing as 25 times 1, so that is 25.

You would get 8 minus 4 is 4.

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Times negative one.

What's that going to simplify to?

So we can do this in multiple ways, simplify different parts.

And you could leave it this way, or you could actually try to simplify this a little bit.

And this number is the same as this number.

We have all of that over minus 16.

Because if we just simplify this expression, and we include things in the simplified domain that weren't included in the original function's domain, we've actually changed the expression.

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We've stated the domain.

This is where we just get our algebra hat on, and try to simplify until it's a separable equation in v.

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Remember, this thing is just this thing right here, we're just doing the Pythagorean theorem.

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let's cancel that out.

Well, a is minus 10, so 2 times a is minus 20.

We have a b squared over b to the third power

You could bring the 13 down here if you want to write it so you don't have a fraction over a fraction.

These will make this expression undefined. It will make this denominator equal to 0.