1. Slide 10- 1 Copyright © 2012 Pearson Education, Inc.

2. Copyright © 2012 Pearson Education, Inc 8.1 Quadratic Equations ■ The Principle of Square Roots ■ Completing the Square ■ Problem Solving

3. Slide 8- 3 Copyright © 2012 Pearson Education, Inc The Principle of Square Roots Let’s consider x 2 = 25. We know that the number 25 has two real-number square roots, 5 and -5, the solutions of the equation. Thus we see that square roots can provide quick solutions for equations of the type x 2 = k.

4. Slide 8- 4 Copyright © 2012 Pearson Education, Inc The Principle of Square Roots For any real number k, If x 2 = k, then

5. Slide 8- 5 Copyright © 2012 Pearson Education, Inc Example Solution Solve 5x 2 = 15. Give exact solutions and approximations to three decimal places. We often use the symbol to represent both solutions. The solutions are which round to 1.732 and –1.732. The check is left to the student. Isolating x 2 Using the principle of square roots

6. Slide 8- 6 Copyright © 2012 Pearson Education, Inc Example Solution Solve 16x 2 + 9 = 0. The solutions are The check is left to the student. Recall that

7. Slide 8- 7 Copyright © 2012 Pearson Education, Inc The Principle of Square Roots (Generalized Form) For any real number k and any algebraic expression X, If X 2 = k, then

8. Slide 8- 8 Copyright © 2012 Pearson Education, Inc Example Solution Solve (x + 3) 2 = 7. The solutions are

9. Slide 8- 9 Copyright © 2012 Pearson Education, Inc Completing the Square Not all quadratic equations can be solved as in the previous examples. By using a method called completing the square, we can use the principle of square roots to solve any quadratic equation.

10. Slide 8- 10 Copyright © 2012 Pearson Education, Inc Example Solution Solve x 2 + 10x + 4 = 0 x 2 + 10x + 25 = –4 + 25 The solutions are Using the principle of square roots Factoring Adding 25 to both sides.

11. Slide 8- 11 Copyright © 2012 Pearson Education, Inc To Solve a Quadratic Equation in x by Completing the Square 1.Isolate the terms with variables on one side of the equation, and arrange them in descending order. 2.Divide both sides by the coefficient of x 2 if that coefficient is not 1. 3.Complete the square by taking half of the coefficient of x and adding its square to both sides.

12. Slide 8- 12 Copyright © 2012 Pearson Education, Inc 4. Express the trinomial as the square of a binomial (factor the trinomial) and simplify the other side. 5. Use the principle of square roots (find the square roots of both sides). 6. Solve for x by adding or subtracting on both sides.

13. Slide 8- 13 Copyright © 2012 Pearson Education, Inc Problem Solving After one year, an amount of money P, invested at 4% per year, is worth 104% of P, or P(1.04). If that amount continues to earn 4% interest per year, after the second year the investment will be worth 104% of P(1.04), or P(1.04) 2. This is called compounding interest since after the first period, interest is earned on both the initial investment and the interest from the first period. Generalizing, we have the following.

14. Slide 8- 14 Copyright © 2012 Pearson Education, Inc The Compound Interest Formula If the amount of money P is invested at interest rate r, compounded annually, then in t years, it will grow to the amount A given by (r is written in decimal notation.)

15. Slide 8- 15 Copyright © 2012 Pearson Education, Inc Example Solution Jackson invested $5800 at an interest rate of r, compounded annually. In 2 yr, it grew to $6765. What was the interest rate? 1.Familiarize. We are already familiar with the compound-interest formula. 6765 = 5800(1 + r) 2 2. Translate. The translation consists of substituting into the formula:

16. Slide 8- 16 Copyright © 2012 Pearson Education, Inc 3. Carry out. Solve for r: 6765/5800 = (1 + r) 2 4. Check. Since the interest rate cannot be negative, we need only to check.08 or 8%. If $5800 were invested at 8% compounded annually, then in 2 yr it would grow to 5800(1.08) 2, or $6765. The number 8% checks. 5. State. The interest rate was 8%.