Question 1877: need help solving the equation by completing the square 3x^2-18x+28=0 Found 2 solutions by usyim88hk, longjonsilver:Answer by usyim88hk(158) (Show Source):
You can put this solution on YOUR website! First move 28 to the other side
3x^2-18x = -28
Then take 3 out from 3x^2-18x
3(x^2-6x) = -28
Complete the square for x^2-6x which should be 9
3(x^2-6x+9) = -28+27 (add 27 because 9 times 3 is 27)
Now factor x^2-6x+9
3(x-3)^2 = -1
Divide -1 by 3
(x-3)^2 = -1/3
This equation has no solution because the square of a number can not be a negative number. You can not take the square root of a negative which is -1/3 in this case. So this problem has no solution.
You can put this solution on YOUR website!
now 2a = 6, so a=3, and a^2 = 9
Adding 9 to both sides, means that the LHS is now the correct form to factorise = +- which has no real solution. Therefore enter Complex numbers... = +-
cheers
Jon.
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