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If the roots of the quadratic equation $\frac12x^2+99x+c=0$ are $x=-99+\sqrt{8001}$ and $x=-99-\sqrt{8001}$, then what is the value of $c$?

Lightning Jun 14, 2018

#3**+1 **

\(12x^2+99x+c=0\)

\(x = -99\pm\sqrt{8001}\)

I think you have written the question wrong. I think both those answers are suppposed to be over 24.

Melody Jun 15, 2018

#4**+2 **

If the roots of the quadratic equation \(\frac12x^2+99x+c=0 \) are \(x=-99+\sqrt{8001}\) and \(x=-99-\sqrt{8001},\) then what is the value of \(c\)?

In the factored form of a quadratic equation: \((x-b)(x-c)=0\) | The two solutions are \(x_1=b;x_2=c\) |

\([x-(-99-\sqrt{8001})][x-(-99+\sqrt{8001})]=0\) | We used the logic above the reach this conclusion. |

\((x+99+\sqrt{8001})(x+99-\sqrt{8001})=0\) | Simplifying |

\(x^2+198x+1800=0\) | Finalizing |

\(\frac12x^2+99x+900=0 \) | Dividing by 2 to get the form the question presents |

\(c=\boxed{900}\)

I hope this helped,

Gavin

GYanggg Jun 15, 2018