Note:
Solve for x in the following equation.
Example 2:
Set the equation equal to zero by subtracting 8x and 15 from both sides of the equation.
Method 1:Factoring
The left side of the equation is not easily factored, so we will not use this method.
Method 2: Completing the square
Add 5 to both sides of the equation
Add to both sides of the equation:
Factor the left side and simplify the right side:
Take the square root of both sides of the equation:
Add to both sides of the equation:
Method 3:Quadratic Formula
The quadratic formula is
In the equation , a is the coefficient of the term, b is the coefficient of the x term, and c is the constant.
Simply insert 1 for a, -1 for b, and -5 for c in the quadratic formula and simplify.
Method 4:
Graphing
Graph and y=0. The graph of y=0 is nothing more than the x-axis. So what you will be looking for is where the graph of crosses the x-axis. Another way of saying this is that the x-intercepts are the solutions to this equation.
You can see from the graph that there are two x-intercepts located at 2.79128784748 and -1.79128784748. This means that there are two real answers: x=2.79128784748 and -1.79128784748.
The approximate answers are 2.79128784748 and -1.79128784748. These answers may or may not be solutions to the original equation. You must check the answers with the original equation. Check these answers in the original equation.
Check the solution x=2.79128784748 by substituting 2.79128784748 in the
original equation for x. If the left side of the equation equals the right
side of the equation after the substitution, you have found the correct
answer.
Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 2.79128784748 for x, then x=2.79128784748 is a solution.
Check the solution x=-1.79128784748 by substituting -1.79128784748 in
the original equation for x. If the left side of the equation equals the
right side of the equation after the substitution, you have found the
correct answer.
The solutions to the equation are -1.79128784748
and 2.79128784748.
Comment:
You can use the solutions to factor the original equation.
For example, since , then
Since , then
Since the product
then we can say that
This means that and are factors of
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