HedwigRaschke95
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(3x+1)(2x-5)=39
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Sagot :

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[tex](3x+1)(2x-5)=39 \\ 6x^2-15x+2x-5=39 \\ 6x^2-13x-5=39 \ \ \ \ \ \ \ \ \ \ |-39 \\ 6x^2-13x-44=0 \\ 6x^2-24x+11x-44=0 \\ 6x(x-4)+11(x-4)=0 \\ (6x+11)(x-4)=0 \\ 6x+11=0 \ \lor \ x-4=0 \\ 6x=-11 \ \lor \ x=4 \\ x=-\frac{11}{6} \ \lor \ x=4 \\ \boxed{x=-1 \frac{5}{6} \hbox{ or } x=4}[/tex]
Panoyin
                       (3x + 1)(2x - 5) = 39
              3x(2x - 5) + 1(2x - 5) = 39
3x(2x) + 3x(-5) + 1(2x) + 1(-5) = 39
          6x² + (-15x) + 2x + (-5) = 39
                  6x² - 15x + 2x - 5 = 39
                          6x² - 13x - 5 = 39
                                        - 39 - 39
                        6x² - 13x - 44 = 0
                                            x = -(-13) ± √((-13)² - 4(6)(-44))
                                                                    2(6)
                                            x = 13 ± √(169 + 1056)
                                                               12
                                            x = 13 ± √(1225)
                                                        12
                                            x = 13 ± 35
                                                      12
                                            x = 13 + 35    ∨    x = 13 - 35
                                                      12                         12
                                            x = 48        ∨        x = -22
                                                  12                         12
                                            x = 4         ∨         x = -1⁵/₆
The solution set us equal to {4, -1⁵/₆}.
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