Asked by salz
Solve the quadratic equation ax2+bx+c=0 using the completing the squares
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Answered by
Reiny
ax2+bx+c=0
ax^2 + bx = -c <--- subtracted c from both sides
x^2 + (b/a)x = -c/a <--- divide everybody by a
x^2 + (b/a)x + b^2/(4a^2) = -c/a + b^2/(4a^2) <--- "completed the square"
(x + b/(2a) )^2 = (b^2 - 4ac)/(4a^2) <---simplified
x + b/(2a) = ±√(b^2 - 4ac)/(2a) <--took √ of both sides
x = -b/(2a) ± √(b^2-4ac)/(2a)
<b>x = ( -b ± √(b^2 - 4ac)/(2a)</b> <-- solved for x and simplified
Beyold the infamous quadratic equation formula
ax^2 + bx = -c <--- subtracted c from both sides
x^2 + (b/a)x = -c/a <--- divide everybody by a
x^2 + (b/a)x + b^2/(4a^2) = -c/a + b^2/(4a^2) <--- "completed the square"
(x + b/(2a) )^2 = (b^2 - 4ac)/(4a^2) <---simplified
x + b/(2a) = ±√(b^2 - 4ac)/(2a) <--took √ of both sides
x = -b/(2a) ± √(b^2-4ac)/(2a)
<b>x = ( -b ± √(b^2 - 4ac)/(2a)</b> <-- solved for x and simplified
Beyold the infamous quadratic equation formula
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