Question

I need help please!!!

Answer 1

x+10/x+3 +90/x^2-30=0

Answer 2

\[\frac{ x+10 }{ x+3 }+\frac{ 90 }{ x^2-30 }=0\]

Answer 3

Are you sure you copied the problem correctly? I especially mean the second denominator?

Answer 4

90/x^2-25

Answer 5

I don't understand your answer. I asked you about a problem and you write a completely different problem.

Answer 6

sorry I am having trouble with the equation editor...the second denominator was wrong, that's why I corrected it.

Answer 7

Oh, I get it.
You mean the correct second denominator is x^2 - 25?

Answer 8

yes

Answer 9

Ok, let's do it.

Answer 10

Okay

Answer 11

The first step is to factor all denominators. Obviously, the left denominator has nothing to factor.
Can you factor the right denominator?

Answer 12

Before we continue, can you please check the rest of the problem is correct?

Is this it?

\(\large \dfrac{ x+10 }{ x+3 }+\dfrac{ 90 }{ x^2-25 }=0\)

Answer 13

yes

Answer 14

Ok.
Remember you changed the second denominator to \(x^2 - 25\)
This is the difference of two squares which factors like this:

\(a^2 - b^2 = (a + b)(a - b) \)

Answer 15

oops.....(x+5)(x-5)=0

Answer 16

Ok. We now have this:

\(\large \dfrac{ x+10 }{ x+3 }+\dfrac{ 90 }{ (x+5)(x-5) }=0\)

Answer 17

yes

Answer 18

Once again, you are positive that the first fraction is correct, right?
The left denominator is definitely x + 3?

Answer 19

yes

Answer 20

Ok.
The next step is to multiply both sides by the LCD of the denominators.

Answer 21

There are three different factors in the denominators, so the LCD is the product of all three factors.

Answer 22

LCD = (x + 3)(x + 5)(x - 5)

Answer 23

Now we multiply both sides by the LCD.

\( (x + 3)(x + 5)(x - 5)\dfrac{ x+10 }{ x+3 }+(x + 3)(x + 5)(x - 5)\dfrac{ 90 }{ (x+5)(x-5)}=\)

\(=(x + 3)(x + 5)(x - 5) \times 0\)

Answer 24

The right side is zero since we are multiplying by zero.


\( \cancel{(x + 3)}(x + 5)(x - 5)\dfrac{ x+10 }{\cancel{ x+3} }+(x + 3)\cancel{(x + 5)}(\cancel{x - 5})\dfrac{ 90 }{ (\cancel{x+5})(\cancel{x-5})}= 0\)


\( (x + 5)(x - 5)(x+10)+ (x + 3) \times 90 = 0\)

\( (x + 5)(x - 5)(x+10)+ 90(x + 3) = 0\)

Now you need to multiply out the three trinomials on the left and 90(x + 3), and combine like terms.

You will end up with a third degree polynomial which needs perhaps trial and error to solve.

What kind of math are you learning?

Answer 25

College Algebra

Answer 26

Have you dealt with solutions to third degree polynomials or only quadratic equations?

Answer 27

Only quadratic equation..this since to be hard

Answer 28

If this problem is in a section that deals with quadratic equations, then there must be a mistake in the problem.