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Question 232333: how do you solve:
1) 3x^2+8x-3<0
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 3x^2+8x-3<0
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You never solve an inequality direcly.
You solve the corresponding equality
and then determine where the solutions
are for the inequlity.
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Solve 3x^2 + 8x -3 = 0
Factor:
3x^2 + 9x - x - 3 = 0
3x(x+3)-(x+3) = 0
(x+3)(3x-1) = 0
x = -3 or x = 1/3
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Draw a number line and mark x = -3 and x = 1/3
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Check a test value in each of the three intervals to see where
the solutions for the inequality are.
(x+3)(3x-1) < 0
Check x = -10; you get -*- > 0 so no solutions in (-inf,-1/3)
Check x = 0 ; you get +*- <0 so solutions in (-3,1/3)
Check x = 10; you get +*+ >0 so no solutions in (1/3,+inf)
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Conclusion: (-3,1/3) contains all the solutions of the inequality.
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Cheers,
Stan H.
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