Chapter 8 Quadratic Equations and Functions § 8.1 The Square Root Property and Completing the Square The Square Root Property The Square Root Property If u is an algebraic expression and d is a nonzero real number, then u 2 d has exactly two solutions: If u 2 d , then u d or u d . Equivalently, If u 2 d , then u d . Blitzer, Intermediate Algebra, 4e – Slide #3 Using the Square Root Property EXAMPLE Solve: 4 x 2 49. SOLUTION To apply the square root property, we need a squared expression by itself on one side of the equation. We can get x 2 by itself if we divide both sides by 4. 4 x 2 49 This is the given equation. x2 49 4 Divide both sides by 4. 49 x 4 49 7 x 3.5 2 4 Apply the square root property. Simplify. Blitzer, Intermediate Algebra, 4e – Slide #4 Using the Square Root Property CONTINUED Check 3.5: Check -3.5: 4 x 2 49 4 x 2 49 43.5 ? 49 4 3.5 ? 49 4 12.25 ? 49 4 12.25 ? 49 2 49 49 true 2 49 49 true The solutions are 3.5 and -3.5. The solution set is {3.5,-3.5}. Blitzer, Intermediate Algebra, 4e – Slide #5 Using the Square Root Property EXAMPLE Solve: 4 x 2 49 0. SOLUTION To solve by the square root property , we isolate the squared expression on one side of the equation. 4 x 2 49 0 This is the given equation. Subtract 49 from both sides. 4 x 2 49 49 2 x 4 x Divide both sides by 4. 49 4 Apply the square root property. Blitzer, Intermediate Algebra, 4e – Slide #6 Using the Square Root Property CONTINUED 49 1 4 49 7 x i i 3.5i 2 4 Check 3.5i: 4 x 2 49 0 x 1 i Check -3.5i: 4 x 2 49 0 43.5i 49 ? 0 2 4 3.5i 49 ? 0 4 12.25i 2 49 ? 0 4 12.25i 2 49 ? 0 2 4 12.251 49 ? 0 4 12.251 49 ? 0 0 0 true 0 0 true The solutions are 3.5i and -3.5i. The solution set is {3.5i,-3.5i}. Blitzer, Intermediate Algebra, 4e – Slide #7 Using the Square Root Property EXAMPLE Solve by the square root property: 3x 22 36. SOLUTION 3x 2 36 2 This is the given equation. Divide both sides by 3. x 22 12 x 2 12 or x 2 12 x 2 12 or x 2 12 x 2 4 3 or x 2 4 3 x 2 2 3 or x 2 2 3 Apply the square root property. Subtract 2 from both sides of each equation. Rewrite radicands. Simplify. Blitzer, Intermediate Algebra, 4e – Slide #8 Using the Square Root Property CONTINUED Check 2 2 3: Check 2 2 3: 3x 2 36 3x 2 36 2 32 3 ? 36 3 2 3 ? 36 2 3 2 2 3 2 36 ? 2 2 2 3 4 3 ? 36 36 36 true 2 3 2 3 ? 36 3 2 3 ? 36 2 3 2 2 3 2 ? 36 2 2 2 3 4 3 ? 36 36 36 true The solutions are 2 2 3. The solution set is 2 2 3 . Blitzer, Intermediate Algebra, 4e – Slide #9 Completing the Square Completing the Square 2 b If x 2 bx is a binomial, then by adding , which is the 2 square of half the coefficient of x, a perfect square trinomial will result. That is, 2 2 b b x bx x . 2 2 2 Blitzer, Intermediate Algebra, 4e – Slide #10 Completing the Square EXAMPLE What term should be added to the binomial so that it becomes a perfect square trinomial? Write and factor the trinomial. x 2 10x SOLUTION x 2 10x 2 10 2 Add 5 . 2 x2 10x 25 x 5 2 Add 25 to complete the square. Blitzer, Intermediate Algebra, 4e – Slide #11 Completing the Square EXAMPLE Solve by completing the square: 2 x 2 5 x 3 0. SOLUTION 2 x 2 5x 3 0 5 3 2 x x 0 2 2 x2 5 3 x 2 2 5 25 3 25 x x 2 16 2 16 2 This is the given equation. Divide both sides by 2. Add 3/2 to both sides. Complete the square: Half of 5/2 is 2 5 5 25 , and . 4 4 16 Blitzer, Intermediate Algebra, 4e – Slide #12 Completing the Square CONTINUED 2 5 24 25 49 x 4 16 16 25 5 49 x 4 25 5 7 x 4 5 7 5 7 5 x or x 5 4 5 4 x 28 25 28 25 or x 20 20 20 20 Factor and simplify. Apply the square root property. 49 49 7 25 25 5 Split into two equations and subtract 5/4 from both sides of both equations. Get common denominators. Blitzer, Intermediate Algebra, 4e – Slide #13 Completing the Square CONTINUED x 3 53 or x 20 20 Simplify. 3 53 3 53 The solutions are and , and the solution set is , . 20 20 20 20 Blitzer, Intermediate Algebra, 4e – Slide #14 Completing the Square in Application A Formula for Compound Interest Suppose that an amount of money, P, is invested at rate r, compounded annually. In t years, the amount, A, or balance, in the account is given by the formula A P1 r . t Blitzer, Intermediate Algebra, 4e – Slide #15 Completing the Square in Application EXAMPLE Use the compound interest formula, A P1 r t , to find the annual interest rate r. In 2 years, an investment of $80,000 grows to $101,250. SOLUTION We are given that P (the amount invested) = $80,000 t (the time of the investment) = 2 years A (the amount, or balance, in the account) = $101,250. Blitzer, Intermediate Algebra, 4e – Slide #16 Completing the Square in Application CONTINUED We are asked to find the annual interest rate, r. We substitute the three given values into the compound interest formula and solve for r. A P1 r t 101,250 80,0001 r 101,250 2 1 r 80,000 81 2 1 r 64 81 81 1 r or 1 r 64 64 2 Use the compound interest formula. Substitute the given values. Divide both sides by 80,000. Simplify the fraction. Apply the square root property. Blitzer, Intermediate Algebra, 4e – Slide #17 Completing the Square in Application CONTINUED 9 9 or 1 r 8 8 9 9 r 1 or r 1 8 8 1 r r 1 17 or r 8 8 81 81 9 64 64 8 Subtract 1 from both sides. Simplify. Because the interest rate cannot be negative, we reject -17/8. Thus, the annual interest rate is 1/8 = 0.125 = 12.5%. We can check this answer using the formula A P1 r . If $80,000 is invested for 2 years at 12.5% interest, compounded annually, the balance in the account is t Blitzer, Intermediate Algebra, 4e – Slide #18 Completing the Square in Application CONTINUED A $80,0001 0.125 $80,0001.125 $101,250. 2 2 Because this is precisely the amount given by the problem’s conditions, the annual interest rate is, indeed, 12.5% compounded annually. Blitzer, Intermediate Algebra, 4e – Slide #19 Isosceles Right Triangles Lengths Within Isosceles Right Triangles The length of the hypotenuse of an isosceles right triangle is the length of a leg times 2. a 2 a a Blitzer, Intermediate Algebra, 4e – Slide #20 Isosceles Right Triangles in Application EXAMPLE A supporting wire is to be attached to the top of a 70-foot antenna. If the wire must be anchored 70 feet from the base of the antenna, what length of wire is required? SOLUTION Since the supporting wire, the antenna, and a line on the ground between the base of the antenna and the base of the supporting wire form an isosceles right triangle as shown below, antenna supporting wire Blitzer, Intermediate Algebra, 4e – Slide #21 Isosceles Right Triangles in Application CONTINUED this implies that the diagram can be represented as follows, using the “lengths within isosceles right triangles” principle. antenna 70 ft. supporting 2 ft. wire 70 70 ft. Therefore the supporting wire must be 70 2 feet which equals about 99 ft. Blitzer, Intermediate Algebra, 4e – Slide #22