## Square roots in quadratic trinomial inequalities

How do we treat expressions under a sqaure root in inequalities ? Like for ex.

x+4< Math.sqrt(-x^2-8x-12) (sorry, using m.physicsforums, so i dont know what to use for a root, so JAVA :p)
I request the use of this very example.
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 So what I like to do is ignore the inequality sign. Treat it as an = sign. Then when I have solutions for the equality I go back and test in the original equation to find the solutions to my inequality. So for your example. Square both sides. (x+4)^2 = -x^2 - 8x -12 Expand x^2 + 8x + 16 = -x^2 - 8x - 12 Get variables on one side and combine like terms 2x^2 + 16x + 28 = 0 Divide by 2 x^2 + 8x + 14 = 0 Solve x = (-8 +/- SQRT(64 - 4*1*14) ) / 2*1 x = (-8 +/- SQRT(64 - 56 ) / 2 x = (-8 +/- SQRT(8) / 2 x = (-8 +/- 2*SQRT(2) / 2 x = -4 +/- 1*SQRT(2) x = -4 +/- SQRT(2) So now that we have our two solutions we want to treat these as critical points and see what happens between them to find what our solution is. So I like to test points. I will test a point greater than both our solutions (0), a point between our two solutinos (-4) and a point less than our two solutions (-10). For x = 0 we get 0+4< Math.sqrt(-0^2-8*0-12) 4 < SQRT(-12) This does not solve our inequality. For x = -4 -4+4< Math.sqrt(-(-4)^2-8(-4)-12) 0 < SQRT(-16 + 32 -12) 0 < SQRT(4) 0 < 2 Success! For x = (-10) (-10)+4< Math.sqrt(-(-10)^2-8(-10)-12) -6 < SQRT(-100 + 80 -12) -6 < SQRT(-32) This does not solve our equation. Now lets just double check our endpoints. x = -4 +/- SQRT(2) Remember that this solution is going to make the original equation EQUAL. That means that since we have strictly less than (<) they will not be solutions, So the solution to the original will be -4 - SQRT(2) < x < -4 + SQRT(2) I hope I have helped.

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