SOLUTION: Please help me answer this: For what values of k will the equation 3x^2=k-2x have real roots?

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Question 948573: Please help me answer this: For what values of k will the equation 3x^2=k-2x have real roots?
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
3x%5E2=k-2x
3x%5E2%2B2x-k=0=> a=3, b=2, and c=-k
The argument of the square root, the expression b%5E2+%96+4ac, is called the "discriminant" because, by using its value, you can discriminate between (tell the differences between) the various solution types.
if b%5E2+%96+4ac%3C0 there is no real roots
if b%5E2+%96+4ac=0 there is one real roots
if b%5E2+%96+4ac%3E0 there are two real roots
so, we need
if b%5E2+%96+4ac=0 there is one real roots
if b%5E2+%96+4ac%3E0 there are two real roots
and I will write both in one as b%5E2+%96+4ac%3E=0
now just plug in a=3, b=2, and c=-k
2%5E2+%96+4%2A3%2A%28-k%29%3E=0
4+%96+12%28-k%29%3E=0
4+%2B12k%3E=0
12k%3E=-4
k%3E=-4%2F12
k%3E=-1%2F3
so, if k=-1%2F3 we will have one real root which is x=-1%2F3 (you can check it)
if k%3E=-1%2F3, using for example k=0 we will have two real roots which are x+=+0 and x=-2%2F3

so, to get real root/roots, you can use any value for k from [-1%2F3,infinity)