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SAT Math 9.2 Algebra and Functions
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SAT Math 9.2 Algebra and Functions

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SAT Math 9.5 Algebra and Functions 213 Views


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Description:

SAT Math 9.5 Algebra and Functions

Language:
English Language

Transcript

00:02

Here's a question that's part of a balanced Shmoopy breakfast.

00:06

If f (x) = x squared – 6x + 9, and  f (x + 1) = 1, what is one possible value of x?

00:16

Well, it wants us to find one possible value of x, which just means… solve for x.

00:21

The requirements for a value of x are: f of x is equal to x squared - 6x + 9, and f of x + 1 = 1.

00:31

We’ll need to put together the two requirements to find x.

00:37

Let’s take a closer look at the second requirement.

00:40

This just means that if we plugged the value (x + 1) into the function,

00:44

which we’re given in the first requirement, we can solve for x.

00:47

To solve this equation, we’re first going to have to turn it back into a polynomial.

00:51

So, we expand (x + 1) squared first.

00:55

By applying foil, we turn (x + 1) squared into x squared + 2x + 1.

01:03

Then, we distribute -6 into the second parentheses. We get -6x - 6.

01:10

Now we can combine like terms. There’s only one x squared term, so that stays by itself.

01:15

However, we have both 2x and -6x, so we can combine those to -4x.

01:20

Then, all of the constants add to +3. Great, we have a polynomial!

01:24

To be more specific… it's a quadratic.

01:27

We can just stick this puppy into the quadratic formula and come out with the answer.

01:31

Plugging in our values, we get that x is equal to 4 plus or minus the square root of 16 – 12 over 2.

01:40

This simplifies to 4 plus or minus 2 over 2. The possibilities are 1 or 3.

01:47

That's our answer either 1 or 3.

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