Add the following expressions and simplify if possible: x 2 x 2 − 4 + 2 x 4 − x 2 \frac{x^2}{x^2-4}+\frac{2x}{4-x^2}

this problem we are asked to add the following expressions and simplify if possible. Now I can see I'm adding two rational expressions. Now what I'm going to do is see how I can factor the numerators and denominators completely of each. This is going to make adding them together a bit more simple, since we're ultimately going to find the least common denominator, or LCD, to solve this problem. Now in the numerator, I can see we have this x squared. And then in the denominator, we have x squared minus four. Four is the same thing as two squared. And notice how I get us to have a difference of two squares right about here. So what I can do is take this whole denominator, x squared of minus two squared, and I can factor that to be x plus two times x minus two. This is the rule for when we have a difference of two squares. So x squared minus four factors to become x plus two times x minus two. Now what I'm going to do is add this to this expression we have on the right. So that's going to be two x over, and then we're going to have four minus x squared. Now what I'm going to do is take this x squared and make it negative. So it's the same thing as four plus negative x squared. But what I can do is flip around the order since we're now adding these, so I can bring this negative x squared to the front, giving us negative x squared plus four. Now what I can do from here is factor out a negative one from this denominator. So we're going to keep everything the same here, except I'm going to factor a negative one out of the denominator. So if I factor a negative one out of the denominator, well that's going to leave me with a negative one on the outside of the parenthesis, or just a negative sign, and then on the inside we'll change the signs of each of these, so that's going to give me x squared, and then we'll have minus four. Now notice x squared minus four is the same thing as x squared minus two squared. This is a difference of squares. So I can go through this process of factoring again. So we'll keep everything the same, but I'm going to use this difference of squares that I've been using in these problems. So this x squared minus two, that's going to become x plus two times x minus two. That's x squared minus two squared. Now we have close common denominators. The main difference that I'm seeing is that we don't have a negative sign over here. Now I could do this a couple different ways. I could multiply the top and bottom by a negative one in this left expression, or what I could do is I could simply bring this negative sign to the front of this expression on the right. And that's what I'm going to do, because a negative sign would just make this whole thing subtracted. So what we're ultimately gonna end up with on top is x squared minus two x, and then what we'll end up with on bottom is x plus two times x minus two. Now what I'm gonna do is factor out this x from each of these terms. We'll have x times x minus two, and then the denominator will have x plus two times x minus two. And I can go ahead and cancel those x minus two's. So all this is going to leave me with is x in the numerator and x plus two in the denominator. And that is how we can solve this problem. Hope you found this video helpful.

Add the following expressions and simplify if possible: x 2 x 2 − 4 + 2 x 4 − x 2 \frac{x^2}{x^2-4}+\frac{2x}{4-x^2}

x 2 ( x − 2 ) ( x + 2 ) \frac{x^2}{\left(x-2\right)\left(x+2\right)} ( x − 2 ) ( x + 2 ) x 2

x x − 2 \frac{x}{x-2} x − 2 x

8. Rational Expressions and Equations

Adding and Subtracting Rational Expressions with Different Denominators

Beginning & Intermediate Algebra

8. Rational Expressions and Equations

Adding and Subtracting Rational Expressions with Different Denominators

Struggling with Beginning & Intermediate Algebra?

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Multiple Choice

Add the following expressions and simplify if possible:
$\frac{x^{2}}{x^{2} - 4} + \frac{2 x}{4 - x^{2}}$

$\frac{x^2}{\left(x-2\right)\left(x+2\right)}$

$\frac{x}{x-2}$

$\frac{1}{x+2}$

$\frac{x}{x+2}$

Verified step by step guidance

Identify the given expression: $\frac{x^2}{x^2 - 4} + \frac{2x}{4 - x^2}$. Notice that the denominators are very similar but not exactly the same.

Factor the denominators to see their relationship: $x^2 - 4$ factors as $(x - 2)(x + 2)$, and \$4 - x^2$ can be rewritten as $-(x^2 - 4)$, so it factors as $-(x - 2)(x + 2)$.

Rewrite the second fraction using this factorization: $\frac{2x}{4 - x^2} = \frac{2x}{-(x - 2)(x + 2)} = -\frac{2x}{(x - 2)(x + 2)}$.

Now both fractions have the same denominator $(x - 2)(x + 2)$, so combine them: $\frac{x^2}{(x - 2)(x + 2)} - \frac{2x}{(x - 2)(x + 2)} = \frac{x^2 - 2x}{(x - 2)(x + 2)}$.

Factor the numerator $x^2 - 2x$ as $x(x - 2)$, then simplify the fraction by canceling the common factor $(x - 2)$ from numerator and denominator, resulting in $\frac{x}{x + 2}$.

6:23m

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Struggling with Beginning & Intermediate Algebra?

Add the following expressions and simplify if possible:x2x2−4+2x4−x2\(\frac{x^2}{x^2-4}\)+\(\frac{2x}{4-x^2}\)

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Add the following expressions and simplify if possible:
$\frac{x^{2}}{x^{2} - 4} + \frac{2 x}{4 - x^{2}}$