#5**+2 **

I believe I have gotten the answer to this riddle! Using the slope formula, one can find the value for a, if a solution exists.

\(m=\frac{y_2-y_1}{x_2-x_1}\)

Just plug in the appropriate values for the x- and y-corrdinates and solve.

\(\frac{5a^2-3a^2}{3a+4-(2a+4)}=a+3\) | Now, solve for a. First, distribute the negation to all the terms inside the parentheses. |

\(\frac{5a^2-3a^2}{3a+4-2a-4}=a+3\) | Simplify the numerator and denominator in the fraction because both happen to have like terms. |

\(\frac{2a^2}{a}=a+3\) | Both the numerator and denominator can cancel out an "a." |

\(2a=a+3\) | Subtract a from both sides. |

\(a=3\) | Wow! There is actually a solution! |

In case you are wondering, this is correct. Click here to view the corresponding graph.

TheXSquaredFactor
Nov 10, 2017

#3**+2 **

I can explain how to simplify for you! First, let's solely worry about the square root of a negative number first.

\(4*\textcolor{blue}{\sqrt{-9}}-2\) | |

\(\sqrt{-9}=\sqrt{9}\sqrt{-1}=\sqrt{9}i=3i\) | By definition, \(i=\sqrt{-1}\). Here, I broke up the radical into two separate parts. |

\(4*\textcolor{blue}{3i}-2\) | Operations with imaginary numbers are the same as with a generic variable. |

\(12i-2\) | Now, rearrange into \(a+bi\) format such that a is the real part and b is the coefficient of the imaginary part. |

\(-2+12i\) | |

TheXSquaredFactor
Nov 10, 2017