- Solve rational equations. - Solve radical equations. [SLIDE 1] Solve rational equations by multiplying both sides of the equation by the LCD, or least common denominator. This will get rid of all of the fractions. [SLIDE 2] `1/(5x+15)-1/(x^2-9) = 5/(x-3)`. So if we start by factoring all of our denominators, we have `1/(5(x+3))-1/((x-3)(x+3)) = 5/(x-3)`. Now we need to multiply both sides by our LCD. `5(x+3)(x-3)`. So now we can start canceling. The `x+3`'s and the 5's cancel on the first fraction, and then the `(x-3)(x+3)`'s' cancel on our second fraction, and the `(x-3)`'s cancel in our third fraction. So that we're left with `x-3-5 = 5*5(x+3)`, so that we have `25(x+3)`, which is `25x+75` on the right and `x-8` on the left. So if we subtract x from both sides, we get 24x on the right, and if we subtract 75 from both sides, this gives us -83. This means that `x = -83/24` is a possible solution. You have to plug back into our x-values in our equation to see if it is a solution. So, `1/(5(-83/24)+15)-1/((-83/24)^2-9) = 5(-83/24-3)`. This gives us `-24/31` on the left, and `-24/31` on the right. `-83/24` checks, so it is a solution. [SLIDE 3] The Principle of Powers: If n is positive and `a = b`, then `a^n = b^n`. [SLIDE 4] Solve the following equation for x. We're given `sqrt(x)+sqrt(4+x) = 3`. So using the Principle of Powers, we can go ahead and square both sides. This will give us `sqrt(x)^2+2(sqrt(x))(sqrt(4+x))+sqrt(4+x)^2 = 9`. This is `x+2sqrt(x(4+x))+4+x = 9`. So `x+2sqrt(4x+x^2)+4+x = 9`. Now let's combine our x's and subtract 4 on both sides, so that we get `2x + 2sqrt(4x+x^2) = 5`. And then subtract 2x from both sides, and we get `2sqrt(4x+x^2) = 5-2x`. And then by the principle of powers, we can square both sides again and get `4(4x+x^2) = 25-2(5)(2x)+4x^2`. This gives us `16x+4x^2 = 25-20x+4x^2`. And now the `4x^2`'s cancel on both sides, and we can add 20x to both sides, so that we get `36x = 25`. `x = 25/36`. So now we need to plug `25/36` back in, to make sure it satisfies our equation. So we have `sqrt(25/36)+sqrt(4+25/36) = 3`. So that gives us `5/6+sqrt(169/36) = 3`, or `5/6+13/6 = 3`. That's `18/6 = 3`, or `3 = 3`, which is a true statement. So that means that `x = 25/36` is a solution.