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Chapter 10 Quadratic Equations and Functions

Chapter 10 Quadratic Equations and Functions. Section 1 Solving Quadratic Equations by Completing the Square. Section 10.1 Objectives. 1 Solve Quadratic Equations Using the Square Root Property 2 Complete the Square in One Variable 3 Solve Quadratic Equations by Completing the Square

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Chapter 10 Quadratic Equations and Functions

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  1. Chapter 10 Quadratic Equations and Functions Section 1 Solving Quadratic Equations by Completing the Square

  2. Section 10.1 Objectives 1 Solve Quadratic Equations Using the Square Root Property 2 Complete the Square in One Variable 3 Solve Quadratic Equations by Completing the Square 4 Solve Problems Using the Pythagorean Theorem

  3. Square Root Property Square Root Property If x2 = p, then x = or x = Example: Solve the equation:

  4. Square Root Property Solving Quadratic Equations Using the Square Root Property Step 1: Isolate the expression containing the square term. Step 2: Use the Square Root Method. Don’t forget the  symbol. Step 3: Isolate the variable, if necessary. Step 4: Check. Verify your solutions.

  5. Square Root Property Example: Solve the equation: Divide each side by 3. Take the square root of each side. Simplify. Check:  

  6. Square Root Property Example: Solve: Take the square root of each side. Simplify. Subtract 3 from both sides. Divide both sides by 2. Continued.

  7. Square Root Property Example continued: Check:  

  8. coefficient of the first-degree term Completing the Square The idea behind completing the square is to “adjust” the left side of a quadratic equation of the form x2 + bx + c in order to make it a perfect square trinomial. Obtaining a Perfect Square Trinomial Step 1: Identify the coefficient of the first-degree term. Step 2: Multiply this coefficient by and then square the result. Step 3: Add this result to both sides of the equation.

  9. Completing the Square Solving a Quadratic Equation by Completing the Square Step 1: Rewrite x2 + bx + c = 0 as x2 + bx = – c by subtracting the constant from both sides of the equation. Step 2: Complete the square in the expression x2 + bx by making it a perfect square trinomial. Whatever you add to the left side of the equation must also be added to the right side. Step 3: Factor the perfect square trinomial on the left side. Step 4: Solve the equation using the Square Root Property. Step 5: Check. Verify your solutions.

  10. Completing the Square Example: Solve: x2 – 6x – 7 = 0 x2 – 6x = 7 Add 7 to both sides. x2 – 6x + 9= 7 + 9 Complete the square in the expression x2 – 6x. (x2 – 3) = 16 Factor the left side. Use the Square Root Property. Simplify. Add 3 to both sides. Continued.

  11. Completing the Square Example continued: Check: x2 – 6x – 7 = 0 x2 – 6x – 7 = 0 72 – 6(7) – 7 = 0 (– 1)2 – 6(– 1) – 7 = 0 49 – 42 – 7 = 0 1 + 6 – 7 = 0 0 = 0  0 = 0 

  12. Complete the square in the expression Completing the Square Example: Solve: 2x2 + 5x – 3 = 0 Add 3 to both sides. 2x2 + 5x = 3 Divide each term by 2. Factor the left side. Continued.

  13. Subtract from both sides. Completing the Square Example continued: Use the Square Root Property. Simplify. Check:  

  14. hypotenuse c b legs a Pythagorean Theorem Pythagorean Theorem In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. c2 = a2 + b2.

  15. 90’ 90’ Pythagorean Theorem Example: A baseball diamond is square. Each side of the square is 90 feet long. How far is it from home plate to second base? Step 1: Identify We want to know how far it is from home plate to second base. Home plate Let c be the distance from home plate to second base. Step 2: Name c Second base Continued.

  16. 90’ c 90’ Pythagorean Theorem Example continued: Step 3: Translate c2 = a2 + b2 Use the Pythagorean Theorem. c2 = 902 + 902 Substitute. c2 = 8100 + 8100 Step 4: Solve c2 = 16200 Continued.

  17. Due to rounding error. Pythagorean Theorem Example continued: 127.3 is not used because length is never negative. Step 5: Check c2 = a2 + b2 127.32 = 902 + 902 16205.29 = 8100 + 8100 16205.29  16200  Step 6: Answer The distance from home plate to second base is approximately 127.3 feet.

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