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

Question 1

Question 2

$\sf(-(-5)\pm√(\sf (-5)^2-4\cdot1\cdot6))/(2\cdot1)$

$\sf(5\pm√(\sf 25-24))/(2)$

$\sf(5\pm√(\sf 1))/(2)$

$\sf(5\pm1)/(2)$

$\sf x'=(5-1)/(2)~~~~~~x''=(5+1)/(2)$

$\sf x'=(4)/(2)~~~~~~~~~~~~x''=(6)/(2)$

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

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Solutions = {2, 3}

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I hope that helps !!

Question 3

x^2-5x+6=0

$x=(-(-5)\pm√((-5)^2-4*1*6))/(2*1)\Leftrightarrow$

$\Leftrightarrow x=(5\pm√(25-4*6))/(2)\Leftrightarrow$

$\Leftrightarrow x=(5\pm√(25-24))/(2)\Leftrightarrow$

$\Leftrightarrow x=(5\pm√(1))/(2)\Leftrightarrow$

$\Leftrightarrow x=(5\pm1)/(2)\Leftrightarrow$

$\Leftrightarrow x=(5-1)/(2)\;\;\;\vee\;\;\;x=(5+1)/(2)\Leftrightarrow$

$\Leftrightarrow x=(4)/(2)\;\;\;\vee\;\;\;x=(6)/(2)\Leftrightarrow$

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

Answer:
$x\in\{2\;;\;3\}$

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