What is (3)x+3+sqrt(x^2+4x-5)>0 ?

The solution to (3)x+3+sqrt(x^2+4x-5)>0 is x>= 1

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(3)x+3+sqrt(x^2+4x-5)>0

(3) x + 3 + x^{2} + 4 x - 5 > 0

x \geq 1

Interval Notation

[1, \infty)

Solution steps

3 x + 3 + x^{2} + 4 x - 5 > 0

Subtract

3 x + 3

from both sides

3 x + 3 + x^{2} + 4 x - 5 - (3 x + 3) > 0 - (3 x + 3)

Simplify

x^{2} + 4 x - 5 > - (3 x + 3)

f (x) > g (x)

then

g (x) < 0 or g (x) \geq 0

For

- (3 x + 3) < 0 : x > - 1

For

- (3 x + 3) \geq 0 :

No Solution

Find singularity points

Find non-negative values for radicals:

x \leq - 5 or x \geq 1

Combine the intervals

(No Solution or x > - 1) and (x \leq - 5 or x \geq 1)

Merge Overlapping Intervals

x \geq 1

Verify Solutions:

x \geq 1

True

The solution is

x \geq 1

x + 2 < x + 5

2 - x > x

\frac{( 2 x - 1 )}{( x - 2 )} < 1

1 - x^{2} + 1 < 3 - x^{2}

8 - 3 x < - 9

What is (3)x+3+sqrt(x^2+4x-5)>0 ?
The solution to (3)x+3+sqrt(x^2+4x-5)>0 is x>= 1