SOLUTION: solve the quadratic equation by completing the square:
3x^2 - 6x + 2 = 0
3x^2-6x = -2
x^2-2x = -2/3
(1/2 of -2 (the x coefficent) squared is 1; add it to both sides)
x^2-2x +
Algebra ->
Quadratic Equations and Parabolas
-> SOLUTION: solve the quadratic equation by completing the square:
3x^2 - 6x + 2 = 0
3x^2-6x = -2
x^2-2x = -2/3
(1/2 of -2 (the x coefficent) squared is 1; add it to both sides)
x^2-2x +
Log On
Question 89375This question is from textbook Beginning Algebra
: solve the quadratic equation by completing the square:
3x^2 - 6x + 2 = 0
3x^2-6x = -2
x^2-2x = -2/3
(1/2 of -2 (the x coefficent) squared is 1; add it to both sides)
x^2-2x + 1 = -2/3 + 1
(x-1)^2 = -5/3
x-1 = +/- radical 5/3
(radical 5/3 = radical 5 over radical 3 multiplied by radical 3 over radical 3)
x-1 = +/- radical 15 over 3
x = 1 +/- radical 15 over 3
but according to the book the answer is 3 +/- radical 3 over 3 - what did i do wrong? This question is from textbook Beginning Algebra
You can put this solution on YOUR website! solve the quadratic equation by completing the square:
3x^2 - 6x + 2 = 0
3x^2-6x = -2
x^2-2x = -2/3
(1/2 of -2 (the x coefficent) squared is 1; add it to both sides)
x^2-2x + 1 = -2/3 + 1
(x-1)^2 = -5/3
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You made an aritmetic error here:
-2/3 + 1 = -2/3 + 3/3 = 1/3 not -5/3
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Cheers,
Stan H.
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