Question

How do you combine `7/(x^2-x-2)+x/(x^2+4x+3)`?

Answer 1

Here's what I got.

Explanation:

Your goal here is to find a common denominator for those two fractions. This will allow you to combine the two numerators and simplify the expression, if possible.

Take a look at the first denominator

`x^2 - x - 2`

If you make this equal to zero, you can use the quadratic formula

`color(purple)(|bar(ul(color(white)(a/a)color(black)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a))color(white)(a/a)|)))`

to find its roots. In this case, you have

`x_(1,2) = (-(-1) +- sqrt( (-1)^2 - 4 * 1 * (-2)))/(2 * 1)`

`x_(1,2) = (1 +- sqrt(9))/2 implies { (x_1 = (1 + 3)/2 = 2), (x_2 = (1 - 3)/2 = -1) :}`

You can thus rewrite the first denominator as

`x^2 - x - 2 = (x-2)(x+1)`

Do the same for the second denominator

`x^2 + 4x + 3`

to get

`x_(1,2) = (-4 +- sqrt( 4^2 - 4 * 1 * 3))/(2 * 1)`

`x_(1,2) = (-4 +- 2)/2 implies {(x_1 = (-4 - 2)/2 = -3), (x_2 = (-4 + 2)/2 = -1) :}`

The second denominator can be rewritten as

`x^2 + 4x + 3 = (x+3)(x+1)`

The expression becomes

`7/((x-2)(x+1)) + x/((x+3)(x+1))`

The common denominator will be

`overbrace((x+1))^(color(purple)("common to both denominators")) * (x-2) * (x+3)`

Multiply the first fraction by `1 = (x+3)/(x+3)` and the second fraction by `1=(x-2)/(x-2)` to get

`7/((x-2)(x+1)) * (x+3)/(x+3) + x/((x+3)(x+1)) * (x-2)/(x-2)`

`(7 * (x+3))/((x+1)(x-2)(x+3)) + (x * (x-2))/((x+1)(x-2)(x+3))`

Now focus on the numerator

`7(x+3) + x(x-2) = 7x+21 + x^2 - 2x`

`=x^2 + 5x + 21`

The expression can thus be rewritten as

`(x^2 + 5x + 21)/((x+1)(x-2)(x+3))`

The numerator cannot be factored in any useful form, i.e. the quadratic

`x^2 + 5x + 21 = 0`

has complex roots, so it's best to leave the answer like this.