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Question 210158: solve 3y^2+4y-2<= 0
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! solve 3y^2+4y-2<= 0
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Solve the EQUALITY:
y = [-4 +- sqrt(16 - 4*3*-2)]/6
y = [-4 +- sqrt(40)]/6
y = [-4 +- 2sqrt(10)]/6
y = (-2/3)+(1/3)sqrt(10) or y = (-2/3)-(1/3)sqrt(10)
Approximate solutions: y = 0.3874 or y = -1.721
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Solve the INEQUALITY:
Draw a number line. Plot the two equality solutions.
Check a value in each of the three intervals formed by those solutions
to see where the solution for the inequality lie.
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3y^2+4y-2<= 0
Check y = -10 ; 3(-10)^2+4(-10)-2 < 0 ; false
Check y = 0 ; -2 < 0 ; true so solutions in (-1.721,0.3874)
check y = 10 ; 3(10)^2-4(10)-2 < 0 ; false
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Final solution; (-2/3)-(1/3)sqrt(10) <= y <= (-2/3)+(1/3)sqrt(10)
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Cheers,
Stan H.
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