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# Quadratic Equations

### Solving Equations by the Quadratic Formula and Quadratic Methods

Question | Answer |
---|---|

What is the Quadratic Equation? | −b±√b^2−4ac/2a |

What form must an equation be in to solve using the Quadratic Method? | ax^2 + bx + c = 0 |

Solve x^2 + 3x – 4 = 0 using the Quadratic Equation | Using a = 1, b = 3, and c = –4, the solutions are x=-4 and x=1 |

Solve 6x^2 + 11x – 35 = 0 | Using a=6, b=11 and c=-35, the solutions are x=-7/2 and x=5/3 |

Solving geometric problems modeled by the Quadratic Equation | 1. Read the problem 2. Assign variables 3. Write the equation 4. Solve 5. Check the solution |

A fenced in a rectangular field has an enclosed area of 75 ft^2. The width of the field is to be 3 feet longer than the length of the field. What are the dimensions of the field? | x= length of the field x+3= width of the field x(x+3)=75 Distribute x : x^2+3x=75 Subtract 75 from both sides: x^2+3x-75=0 Solve using the Quadratic Equation x=7.29 and x= -10.29 Disregard the negative solution |

What does it mean for an equation to be reducible to Quadratic form? | It means to make an equation look like a quadratic equation |

Solve x^4 − 20x^2 + 64 = 0 | Let u = x^2 u^2 − 20u + 64 = 0 Factor: (u − 16)(u − 4) = 0 So the solutions for u are 16 and 4. |

An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t^2 + 19.6t + 58.8, where s is in meters. When does the object strike the ground? | Let s = 0 and set s equal to zero, and solve: 0 = –4.9t^2 + 19.6t + 58.8 0 = t^2 – 4t – 12 0 = (t – 6)(t + 2) Then t = 6 or t = –2 *So "t = –2" is an extraneous solution, and will be ignored* |

One number is the square of another. Their sum is 132. Find the numbers. | Let A and B be the numbers. A = B^2 The sum is 132, so A + B = 132 A = B^2 into A + B = 132 and solve for B: The possible solutions are B = -12 and B = 11 |

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