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###### Exercises

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Exercises 1 | |

Exercises 2 | |

Exercises 3 | |

Exercises 4 | |

Exercises 5 | |

Exercises 6 | |

Exercises 7 | |

Exercises 8 | |

Exercises 9 | |

Exercises 10 | |

Exercises 11 | |

Exercises 12 | |

Exercises 13 | |

Exercises 14 | |

Exercises 15 | |

Exercises 16 | |

Exercises 17 | |

Exercises 18 | |

Exercises 19 | |

Exercises 20 | |

Exercises 21 | |

Exercises 22 | |

Exercises 23 Let's recall the formula for the volume of a rectangular prism. V=ℓwh Here ℓ is the length of the prism, w is the width of the prism, and h is its height. In our case V=380 and h=5. We also know that the length ℓ is four times the width w, so ℓ=4w. We will substitute these values into the formula. By doing so, we will obtain an equation that can be solved for w. V=ℓwhSubstitute V=380,h=5,ℓ=4w380=4w(w)(5)Multiply380=20w2LHS/20=RHS/2019=w2LHS=RHS±19=wRearrange equationw=±19 The solutions to the equation are 19 and -19. Since w represents the width of the prism, it cannot be negative. Therefore, w=19 and ℓ=419. We can use a calculator to approximate our results. wℓ=19≈4.4=419≈17.4 The width is 19≈4.4 meters and the length is 419≈17.4 meters. | |

Exercises 24 aWe are given an equation that models the height h (above the water) of the fishing lure after t seconds. h=-16t2+24t+4 To find how long it takes for the fishing lure to reach a height of 12 feet, let's substitute h=12 into our equation. 12=-16t2+24t+4 We will solve this equation by graphing. There are three steps to solving a quadratic equation by graphing.Write the equation in standard form, at2+bt+c=0. Graph the related function f(t)=at2+bt+c. Find the t-intercepts of the graph, if any.The solutions of at2+bt+c=0 are the t-intercepts of the graph of f(t)=at2+bt+c. Let's get started!Writing the Equation in Standard Form We will rewrite our equation in standard form. This means gathering all of the terms on the left-hand side of the equation. 12=-16t2+24t+4LHS−12=RHS−120=-16t2+24t−8Rearrange equation-16t2+24t−8=0 To make the next steps easier, let's simplify the equation as much as possible. First we will multiply both sides of the equation by -1. -16t2+24t−8=0LHS⋅(-1)=RHS⋅(-1)(-16t2+24t−8)⋅(-1)=0Distribute (-1)16t2−24t+8=0 Next we will factor out the greatest common factor. To find it, let's factor each term and determine the common factors. 16t224t8=2⋅2⋅2⋅2⋅t⋅t=2⋅2⋅2⋅3⋅t=2⋅2⋅2 The greatest common factor of the three terms is 2⋅2⋅2=8. Let's factor it out! 16t2−24t+8=0Rewrite 16t2 as 8⋅2t28⋅2t2−24t+8=0Rewrite -24t as 8⋅(-3t)8⋅2t2+8⋅(-3t)+8=0Rewrite 8 as 8⋅18⋅2t2+8⋅(-3t)+8⋅1=0Factor out 88(2t2−3t+1)=0LHS/8=RHS/82t2−3t+1=0Graphing the Related Function Let's write the function related to our equation. Equation: 2t2−3t+1=0Finding the x-intercepts of the Graphb5y+20 |

##### Other subchapters in Solving Quadratic Equations

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Properties of Radicals
- Solving Quadratic Equations by Graphing
- Solving Quadratic Equations Using Square Roots
- Solving Quadratic Equations by Completing the Square
- Solving Quadratic Equations Using the Quadratic Formula
- Solving Nonlinear Systems of Equations
- Chapter Review
- Chapter Test
- Cumulative Assessment