Exercise 48 Page 196

We found that the solutions to the equation are x=- 5 and x=3.

Solving the Equation by Completing the Square

To complete the square we can follow the steps shown below.

1. Rewrite the equation in the form x^2+bx = c.
2. Calculate half of the coefficient of the linear term. The result can be expressed as b2.
3. Calculate the square of b2.
4. Add ( b2)^2 to both sides of the equation.
5. The left-hand side is now a perfect square trinomial, and can be rewritten as (x+ b2)^2.

Our equation can be rewritten as indicated in the first step. x^2-2x-15 = 0 ⇒ x^2 + (-2)x = 15 We can see that b=- 2. With this information, we can calculate the value of b2. b=- 2 ⇔ b/2=- 1 We can now calculate the value of ( b2)^2.

Now, we will add the value of ( b2)^2, which is 1, to both sides of the equation x^2-2x=15. Then, we will factor the resulting perfect square trinomial and solve the equation.

The solutions are 5 and - 3.

Solving the Equation Using the Quadratic Formula

We can use the Quadratic Formula to solve any quadratic function in standard form. Let's recall the Quadratic Formula. x = - b ± sqrt(b^2-4ac)/2a To use it, we need to identify the values of the a, b, and c. Standard Form:& ax^2+ bx+ c=0 Given Equation:& 1x^2+( - 2)x + ( -15) = 0 We see that a= 1, b= - 2, and c= - 15. Now we can use the Quadratic Formula to solve our equation.

The solutions are 5 and - 3.

Method Comparison

Since the coefficient of the linear term of the equation is even, completing the square is the most practical method. Factoring is also a good option for this case, since the coefficients are integers and factors can be easily determined. On the other hand, using the Quadratic Formula may not be practical due to the possibility of making mistakes during calculations.