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**Chapter 8 Quadratic Equations and Functions**

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§ 8.1 The Square Root Property and Completing the Square

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**Introduction What we already know about quadratic equations…**

A quadratic equation can be written in the standard form: Some quadratic equations can be solved by factoring. . Some quadratic equations cannot be factored. In this section, we look at a method for solving a quadratic equation that won’t factor. We look at a method called Completing the Square. Blitzer, Intermediate Algebra, 5e – Slide #3 Section 8.1

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**The Square Root Property**

If u is an algebraic expression and d is a nonzero real number, then has exactly two solutions: Equivalently, This property says that when we take the square root of both sides of an equation, we get two roots. Don’t forget the Blitzer, Intermediate Algebra, 5e – Slide #4 Section 8.1

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**Using the Square Root Property**

EXAMPLE Solve: SOLUTION To apply the square root property, we need a squared expression by itself on one side of the equation. We can get by itself if we divide both sides by 4. This is the given equation. Divide both sides by 4. Apply the square root property. Simplify. Blitzer, Intermediate Algebra, 5e – Slide #5 Section 8.1

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**Using the Square Root Property**

CONTINUED Check 3.5: Check -3.5: ? ? ? ? true true The solutions are 3.5 and The solution set is {3.5,-3.5}. Blitzer, Intermediate Algebra, 5e – Slide #6 Section 8.1

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**Using the Square Root Property**

EXAMPLE Solve: SOLUTION To solve by the square root property , we isolate the squared expression on one side of the equation. This is the given equation. Subtract 49 from both sides. Divide both sides by 4. Apply the square root property. Blitzer, Intermediate Algebra, 5e – Slide #7 Section 8.1

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**Using the Square Root Property**

CONTINUED Check 3.5i: Check -3.5i: ? ? ? ? ? ? true true The solutions are 3.5i and -3.5i. The solution set is {3.5i,-3.5i}. Blitzer, Intermediate Algebra, 5e – Slide #8 Section 8.1

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**Using the Square Root Property**

EXAMPLE Solve by the square root property: SOLUTION This is the given equation. Divide both sides by 3. Apply the square root property. Subtract 2 from both sides of each equation. Rewrite radicands. Simplify. Blitzer, Intermediate Algebra, 5e – Slide #9 Section 8.1

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**Using the Square Root Property**

CONTINUED Check : Check : ? ? ? ? ? ? ? ? true true The solutions are The solution set is Blitzer, Intermediate Algebra, 5e – Slide #10 Section 8.1

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**Completing the Square Completing the Square**

If is a binomial, then by adding , which is the square of half the coefficient of x, a perfect square trinomial will result. That is, Blitzer, Intermediate Algebra, 5e – Slide #11 Section 8.1

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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. SOLUTION Add Add 25 to complete the square. Blitzer, Intermediate Algebra, 5e – Slide #12 Section 8.1

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**Completing the Square Solve by completing the square:**

EXAMPLE Solve by completing the square: SOLUTION This is the given equation. Divide both sides by 2. Add 3/2 to both sides. Complete the square: Half of 5/2 is Blitzer, Intermediate Algebra, 5e – Slide #13 Section 8.1

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**Completing the Square Factor and simplify.**

CONTINUED Factor and simplify. Apply the square root property. Split into two equations and subtract 5/4 from both sides of both equations. Get common denominators. Blitzer, Intermediate Algebra, 5e – Slide #14 Section 8.1

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**Completing the Square Simplify.**

CONTINUED Simplify. The solutions are , and the solution set is Blitzer, Intermediate Algebra, 5e – Slide #15 Section 8.1

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**A Formula for Compound Interest**

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 Blitzer, Intermediate Algebra, 5e – Slide #16 Section 8.1

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**Completing the Square in Application**

EXAMPLE Use the compound interest formula, , 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, 5e – Slide #17 Section 8.1

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**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. 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, 5e – Slide #18 Section 8.1

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**Completing the Square in Application**

CONTINUED 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 = = 12.5%. We can check this answer using the formula If $80,000 is invested for 2 years at 12.5% interest, compounded annually, the balance in the account is Blitzer, Intermediate Algebra, 5e – Slide #19 Section 8.1

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**Completing the Square in Application**

CONTINUED 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, 5e – Slide #20 Section 8.1

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**Lengths Within Isosceles Right Triangles**

The length of the hypotenuse of an isosceles right triangle is the length of a leg times a a a Blitzer, Intermediate Algebra, 5e – Slide #21 Section 8.1

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**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, 5e – Slide #22 Section 8.1

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**Isosceles Right Triangles in Application**

CONTINUED this implies that the diagram can be represented as follows, using the “lengths within isosceles right triangles” principle. 70 ft. antenna supporting wire ft. 70 ft. Therefore the supporting wire must be feet which equals about 99 ft. Blitzer, Intermediate Algebra, 5e – Slide #23 Section 8.1

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**In Summary… Solving Quadratic Equations by Completing the Square**

If the coefficient of the second degree term is not 1, divide both sides by this coefficient Isolate variable terms on one side Complete the square by adding the square of half the coefficient of x to both sides Factor the perfect square trinomial Solve by applying the square root property Blitzer, Intermediate Algebra, 5e – Slide #24 Section 8.1

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