Solve equations reducible to quadratics
Example 1.
![Picture](/uploads/1/1/5/3/11531538/8327669.png)
In this equation, we can view (x squared + 2x) as a pronumeral for example y. Therefore, the equation becomes y squared - 4y + 3 = 0. We solve this to get (y - 3) (y - 1) = 0. Therefore y = 3 or y = 1. We then substitute (x squared + 2x) to get:
x squared + 2x = 3 or
x squared + 2x = 1. These are then viewed as 2 separate quadratic equations. We will look at the first one.
![Picture](/uploads/1/1/5/3/11531538/1814772.png)
This is easily factorised to get: (x + 3) (x - 1) = 0. Therefore x = -3 or x = 1
The second equation is shown on the left. The quadratic equation is then used to get the following solutions:
The second equation is shown on the left. The quadratic equation is then used to get the following solutions: