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AlexFokin [52]
3 years ago
8

Solve the following equation, using the Bhaskara formula: x² - 5x + 6 = 0

Mathematics
2 answers:
kramer3 years ago
4 0

    x^2-5x+6=0

    x=\frac{-(-5)\pm\sqrt{(-5)^2-4\times1\times6}}{2\times1}\Leftrightarrow

\Leftrightarrow x=\frac{5\pm\sqrt{25-4\times6}}{2}\Leftrightarrow

\Leftrightarrow x=\frac{5\pm\sqrt{25-24}}{2}\Leftrightarrow

\Leftrightarrow x=\frac{5\pm\sqrt{1}}{2}\Leftrightarrow

\Leftrightarrow x=\frac{5\pm1}{2}\Leftrightarrow

\Leftrightarrow x=\frac{5-1}{2}\;\;\;\vee\;\;\;x=\frac{5+1}{2}\Leftrightarrow

\Leftrightarrow x=\frac{4}{2}\;\;\;\vee\;\;\;x=\frac{6}{2}\Leftrightarrow

\Leftrightarrow x=2\;\;\;\vee\;\;\;x=3

Answer:  x\in\{2\;;\;3\}

babymother [125]3 years ago
4 0

\sf\dfrac{-(-5)\pm\sqrt{\sf (-5)^2-4\cdot1\cdot6}}{2\cdot1}

\sf\dfrac{5\pm\sqrt{\sf 25-24}}{2}

\sf\dfrac{5\pm\sqrt{\sf 1}}{2}

\sf\dfrac{5\pm1}{2}

\sf x'=\dfrac{5-1}{2}~~~~~~x''=\dfrac{5+1}{2}

\sf x'=\dfrac{4}{2}~~~~~~~~~~~~x''=\dfrac{6}{2}

\sf x'=2~~~~~~~~~~~~~x''=3

⠀⠀

Solutions = {2, 3}

⠀⠀

⠀⠀

I hope that helps !!

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