For what value of the constant k\(\) does the quadratic \( 2x^2 - 5x + k\) have a double root?

Guest Dec 27, 2017

edited by
Guest
Dec 27, 2017

#3**+2 **

Rahuan, I am not completely sure what it means either, but I will interpret "double root" as a zero with a multiplicity of two. Assuming the above is true, we can use the property of the discriminant of a quadratic, \(b^2-4ac\) . The discriminant, if equal to zero, results in a quadratic with a "double root."

\(b^2-4ac=0\) | Plug in the given values of a, b, and c of the quadratic. |

\((-5)^2-4(2)(k)=0\) | Now, solve for k. |

\(25-8k=0\) | Subtract 25 from both sides. |

\(-8k=-25\) | |

\(k=\frac{25}{8} \) | |

TheXSquaredFactor
Dec 27, 2017

#3**+2 **

Best Answer

Rahuan, I am not completely sure what it means either, but I will interpret "double root" as a zero with a multiplicity of two. Assuming the above is true, we can use the property of the discriminant of a quadratic, \(b^2-4ac\) . The discriminant, if equal to zero, results in a quadratic with a "double root."

\(b^2-4ac=0\) | Plug in the given values of a, b, and c of the quadratic. |

\((-5)^2-4(2)(k)=0\) | Now, solve for k. |

\(25-8k=0\) | Subtract 25 from both sides. |

\(-8k=-25\) | |

\(k=\frac{25}{8} \) | |

TheXSquaredFactor
Dec 27, 2017

#4**0 **

does he mean like finding the value of k and then putting the answer as \(\frac{5^2}{8}\)

Rauhan
Dec 27, 2017

#6**0 **

I do not believe so; there is no reason not to simplify the fraction completely.

TheXSquaredFactor
Dec 28, 2017