Question

Algebra Help:
Find the roots of each equation by factoring
12x=9x^2+4
16x^2=9

Answer 1

To factor an expression, you first want all the terms on one side of the '=' sign:
\[ 0 = 9x^2-12x+4 \]
Now, you're looking for a factoring that looks roughly like
\[ (ax\pm b)(cx \pm d)\]
So, what numbers multiply together to get 9 and what other numbers multiply together to get 4?

Answer 2

3 and 3

Answer 3

2 and 2

Answer 4

Great! One more thing, when looking at expressions like
\[ 9x^2 -12x+4\]
the second operation tells us whether the pluses and minuses in
\[(ax\pm b)(cx \pm d)\]
will be the same or different.

We have a '+', so we know the signs are going to be the same. And then '-' before 12x tells us that both those signs are going to be minuses.

Test it out! Does
\[ (3x-2)(3x-2) = 9x^2 -12 + 4\]
?

Answer 5

Yeah it does.

Answer 6

Would that be the root?

Answer 7

Cool. The roots of the equation are the values of x that make the function 0.

So, we've turned
\[ 9x^2-12x+4 = 0 \]
into
\[(3x-2)(3x-2) = 0 \]

For what value of x does the above equation hold true?

Answer 8

1?

Answer 9

Not quite. When x = 1,
\[ (3x-2)(3x-2) = (3-2)(3-2) = 1 \]

Think of it this way: you're multiplying two numbers and getting a 0. That means one of those numbers has to be equal to zero, right?

So, how do you solve 3x-2= 0?

Answer 10

.66

Answer 11

thats what I got.

Answer 12

Correct! So 2/3 is a root of 12x= 9x^2-4

Answer 13

Is there 2 roots?

Answer 14

Or just the one?

Answer 15

There are two roots, but in this case both roots happen to have the value 2/3.

Answer 16

If you had two different numbers. Would you do it with both of the equations?

Answer 17

Also. How do I do the second one?

Answer 18

I'm not sure I understand your first question. If you had two different roots (say a and b), your equation would look something like
\[ (x-a)(x-b) = 0 \]

Try repeating the same process for the second one. Put all the x's and numbers on one side of the equal sign. And then think of multiples

Answer 19

Okay. Thats what I meant. Can you help me factor the second one?

Answer 20

Glad I understood your question then haha

Sure! So first, rewrite the equation into
\[ 16x^2 - 9 = 0 \]
Again, you're looking for ways to break up 16 and 9

Answer 21

4 and 4 for sixteen and 3 and 3 for nine
So you find the x for both roots?

Answer 22

both equations

Answer 23

Which equations are you talking about?

Answer 24

so if you factor 16 and 9, you get
(x+4)^2=0 and (x+3)^3=0 right?

Answer 25

Not quite, beacause
\[ (x+4)(x+3) \neq 16x^2-9\]

Remember that the factors you find for 16 go into the a/c positions and the factors of 9 go into the b/d positions in:
\[ (ax\pm b)(cx \pm d)\]

Answer 26

oh. Okay so it would be -3/4

Answer 27

That's one of the roots! Now the second one...

Answer 28

Second one?

Answer 29

The equation has 2 roots which are different.
When you factored, you should have gotten
\[ (4x-3)(4x+3) \]

Answer 30

the other would be 3/4

Answer 31

Can you help with another thing?

Answer 32

Yup :)

Answer 33

Sure!

Answer 34

Write a quadratic function in standard form for each given set of zeros.
-2 and 7
1 and -8

Answer 35

Ah, so it's the same process, but in reverse. You're given the roots and need the equation.

For -2, you're looking for some expression involving an x so that when x = -2, the expression will equal 0. Does that make sense?

Answer 36

Not really.

Answer 37

Haha okay. Let me use your first question as an example.

When we were looking for roots, we had \[ 3x-2 = 0 \]right?

And we used that to solve for x and get x=2/3.
From this equation, you can also work backwards to the first!
x=2/3
3x = 2
(3x - 2) = 0

Does that make sense?

Answer 38

Yeah. But I dont know how to get the other two numbers.

Answer 39

What 2 numbers are you talking about?

Now you're given -2 and 7.
Let's look at the 7. This means there's a root at x=7.
x=7
(x-7) = 0

If you do the same for x=-2, you get (x+2) = 0.

So your expression looks like
(x+2)(x-7) = 0

Answer 40

Oh. WHat do I do next? Or is that it?

Answer 41

Well, the problem statement says "quadratic function in standard form" so you're probably supposed to multiply the two terms together and get something like x^2+_x+_ = 0

Answer 42

(by probably, I mean you are xD)

Answer 43

Can you help with the other one? I got x^2-5x-14

Answer 44

For the one we just did

Answer 45

You're correct about the first one!

Why don't you take the lead on this question and I'll tell you how you're doing? It's very similar to the one we just did.
You're starting off with x=1 and x=-8.

Answer 46

(x+1)(x-8)=0

Answer 47

Very close. You've got your signs switched for both!
Remember that x=-8 ---> x+8 = 0

Answer 48

Oh. Then you would get x^2+7x-8

Answer 49

Yay! Yes, you're right.

Answer 50

Okkay. Thank you. That's all I needed help with.

Answer 51

You're very welcome. I'm happy to help :)