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# Quadratic Equat.1

### Solving equations by the quadratic formula and quadratic methods

Question | Answer |
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By knowing this value, we can discriminate among the possible number and type of solutions of a quadratic equation. | Discriminant |

What are the steps to solving a quadratic equation? | 1. If the equation is in the form (ax+b)^2, use the square root property and solve. If not, write the equation in standard form (ax^2+bx+c=0) 2.Try to solve the equation by the factoring method. If not possible, use quadratic formula |

Solve p^4-3p^2-4=0 | (p^2-4)(p^2+1)=0 (p-2)(p+2)(p^2+1)=0 p-2=0 or p+2=0 or p^2+1=0 p=2; p=-2; p=+-i (the square root of -1) |

Define quadratic equation. | A second degree equation which is written as Ax^2+Bx+C=0 |

Solve this equation: m^2-m-2=0 | m^2-m-2=0 (m-2)(m+1)=0 m-2=0 or m+1=0 m=2, m=-1 |

Use the quadratic formula to solve this equation: x^2+7x+4=0 | -b=-7; (b^2-4ac)=(7^2-4(1)(4); 2a=(2)(1) (-7+-√49-16)/2 (-7+-√33)/2 Solutions: (-7+√33)/2 and (-7-√33)/2 |

Use the quadratic formula to solve this equation: x^2+3x=1 | x^2+3x-1=0 (-(3)+-√(3)^2-(4)(1)(-1))/(2)(1) (-3+-√9-(-4))/2 Solutions: (-3+√13)/2 and (-3-√13)/2 |

Use the quadratic formula to solve this equation: (x+5)(x=3)=5 | Multiply the binomials: x^2+8x+15-5=5-5 x^2+8x+10=0 (-(8)+-√(8)^2-4(1)(10))/(2)(1) (-8+-√64-40)/2 (-8+-√24)/2=(-8+-2√6)/2 (-8+2√6)/2 or (-8-2√6)/2 Solutions: -4+√6 and -4-√6 |

If b^2-4ac is positive, | There will be two real solutions. |

If b^2-4ac is negative, | There will be two complex solutions, but they will not be real. |

If b^2-4ac is zero, | There will be one real solution. |

Created by:
kaylaroszkowski