# Square roots in quadratic trinomial inequalities

1. Jun 20, 2012

### Kartik.

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.

2. Jun 20, 2012

### Diffy

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.

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.