SOLUTION: solve the equation in real # system x^3-3/4x^2+31/4x+2=0 real solutions of the equation are x=? dont see why my tutor i was assigned cant do these, please help please sho

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Question 641616: solve the equation in real # system
x^3-3/4x^2+31/4x+2=0
real solutions of the equation are x=?
dont see why my tutor i was assigned cant do these, please help
please show work
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Multiply EVERY number by the LCD 4 to go from

x^3-3/4x^2+31/4x+2=0

to

4x^3-3x^2+31x+8=0


Now use the rational root theorem to list out all of the possible rational roots.

The list of all possible rational roots are



It turns out that x=-1%2F4 is the only rational root. You find this out by plugging in each possible rational root and seeing if the result is 0. If it is 0, then you have a rational root.

So x%2B1%2F4 or 4x%2B1 is a factor. Use polynomial long division to get



Now set the quotient x%5E2+-+x+%2B+8 equal to zero and solve for x (using the quadratic formula)


x%5E2-x%2B8+=+0

a = 1, b = -1, c = 8

x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x+=+%28-%28-1%29%2B-sqrt%28%28-1%29%5E2-4%281%29%288%29%29%29%2F%282%281%29%29

x+=+%281%2B-sqrt%281-%2832%29%29%29%2F%282%29

x+=+%281%2B-sqrt%28-31%29%29%2F2

That last line has a negative radicand, which will lead to 2 imaginary or complex solutions (which are NOT real solutions)

So there is only one real solution and it is x=-1%2F4