click below

click below

Normal Size Small Size show me how

# module 17

### Solving Quadratic Equations by the Quadratic Functions

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

solve by using quadratic formula x^2+3x+2=0 | x= 1 or x= -2 |

solve by using quadratic formula (x+3)(x+2)=2 | x= -1 or x= - 4 |

How many solution(s) will a negative discriminant have? | two complex but not real solutions |

If the discriminant is positive how many solution(s) will it have? | two real solutions |

If the discriminant is zero how many solutions will it have? | one real solution |

a=1 b=3 c=4 use a, b, and c to find the discriminant then determine how many solution(s) it will have | -13, negative= two complex but not real solutions |

a=2 b=20 c=1 use a, b, and c to find the discriminant, then determine how many solution(s) it will have | 12, positive= two real solutions |

use quadratic formula to solve (x+6)(x-1)=-2 | x=-5+/- square root 41 all over 2 |

a=1 b=4 c=1 use a, b, and c to find the discriminant then determine how many solution(s) it will have. | x=0, zero= one real solution |

2/3x^2+4/3x+1/3=0 | x= -1+/- 1/2 sq root 2 |