Question

Which real numbers cannot be used in place of the variable? 3b+1/b2-3b-4

Answer 1

Can you write that equation using the equation editor please.

Answer 2

Where is the editor? I'm new to this

Answer 3

In the box you type in it should be just below on the left. You'll see the "Equation" button.

Answer 4

I don't see it. It's a fraction. 3b +1 over b2-3b-4

Answer 5

Okay, so in math there is a rule where you cannot divide by zero meaning the denominator cannot be zero. If it is it would be undefined. Thus you're looking for when the denominator would be zero basically.

Answer 6

Assuming this is the correct equation:\[\frac{3b+1}{b^2-3b-4}\]You'd need to set the bottom equal to zero: `b^2-3b-4=0` and solve for b.

Answer 7

So I got b(b-3)=4

Answer 8

Are you familiar with the quadratic formula?

Answer 9

For equations like:\[ax^2+bx+c=0\]you can use the following to solve for the variable:\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 10

ok

Answer 11

If you do this you will find that b = -1 and 4.

Answer 12

Ok

Answer 13

Do you understand?

Answer 14

I get -7/2

Answer 15

Well you will get two answers because there is a b^2 in the problem but what do you get for your a, b, and c constants to plug into the quadratic equation?

Answer 16

A=1 squared b=-3 c=-4

Answer 17

Okay so plugging them into the equation:\[\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]you get:\[\frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-4)}}{2(1)}\]right?

Answer 18

Now I get it. So now how do I plug in 3b + 1?

Answer 19

The numerator is irrelevant actually. It can be any number and it wouldn't matter. The only thing to worry about is if the bottom becomes zero you will get an error and that only occurs when `b=-1 or 4`.

Answer 20

So do I solve it and get the root of 25

Answer 21

Using quadratic formula is solving this problem. The answers are `-1 and 4`. That's all there is to it.

Answer 22

Oh wow thank u!

Answer 23

You're welcome!