how do i solve 3x^2+x-2=0 by completing the sqaure?
idk, haha. My teacher doesnt explain how do these things very well. the homework says, Solve each equation by completing the square. \[3x ^{2}+x-2=0\]
& thats just one of them
alright well first you'd want to add 2 to the other side and divide by 3
then that'd leave \[x ^{2}+x=0\]
dont you add 2 to both sides of the equation?
add 2 to both sides... i mean't you want to put the 2 on the right side
so you should get \[x^2+\frac{1}{3}x=\frac{2}{3}\]
then take your middle coefficient divide it by two and square it you should get 1/36 but remember you must add to both sides of the equation to keep the equality true \[x+\frac{1}{3}x+\frac{1}{36}=\frac{2}{3}+\frac{1}{36}\]
ohhhhh then ?
now you have a perfect square on the left side so you can factor it \[(x+\frac{1}{6})^2=\frac{2}{3}+\frac{1}{36}=\frac{25}{36}\]
now to solve for x, what would you want to do next?
still there?
hmmmmm..
hint* you can't solve for x when squared
idk D:
haha =] alright well we have so far this, \[(x+\frac{1}{6})^2=\frac{25}{36}\] in order to solve for x we need to get rid of the squared because we do not like it squared, because how would we be able to find the roots as is. so what we need to do next is take the square root of both sides. doing this what do you get?
\[25\div36 \] how do we unsquare that?
\[\sqrt{(x+\frac{1}{6})^2}=\sqrt{\frac{25}{36}}\] is this what you got so are you asking how we'd get rid of the square root?
yes
well using the square root rule that you should have learned and gone over \[\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\] for all values >0
now you should be able to take the square root easily of the top and bottom
oh yes
ok well ithink igot it. thankyou !
np don't forget the positive and negative... you should get the answers to the factors (3x-2)(x+1) where x=-1,2/3
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