Prove that the inequality 3x^2 + 13 < 12x has no real solution.?

Question

Please can somebody help me with this, thankyou very much =]

Answer 1

setting up standard form:

3x² - 12x + 13 = 0
x = 12/6 ± √(144 - 12(39) / 6

note the 144 - 12(39) = -324, the discriminant, is negative, which means no real roots to the equation, which means the opening upward parabola doesn't cross the x axis, so no part of it is below the x axis, so the inequality will have no solutions.

Answer 2

3x^2-12x+13<0
quadratic formula:
x=(-b±√(b^2-4ac))/2a
x= (12±√(144-156))/6
x= (12±√(-12))/6
x= 2±(√12i)/6

X= 2+ 2√3i, 2-2√3i
not a real solution since we have a negative root.

Answer 3

i am pretty sure u have to differentiate them, but i can't quite remember how to do this - i last did it over a year ago. maybe try http://www.physicsforums.com/showthread.php?t=174649 for help