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

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

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

(- 5 + \frac{2}{3} 5 + 4 x - x^{2}) < 0

- 1 \leq x \leq 5

Interval Notation

[- 1, 5]

Solution steps

- 5 + \frac{2}{3} 5 + 4 x - x^{2} < 0

Move

5

to the right side

\frac{2}{3} 5 + 4 x - x^{2} < 5

Multiply both sides by

3

2 5 + 4 x - x^{2} < 15

Divide both sides by

2

5 + 4 x - x^{2} < \frac{15}{2}

Square both sides

(5 + 4 x - x^{2})^{2} < (\frac{15}{2})^{2}

Simplify

5 + 4 x - x^{2} < (\frac{15}{2})^{2}

Expand

(\frac{15}{2})^{2} : \frac{225}{4}

5 + 4 x - x^{2} < \frac{225}{4}

5 + 4 x - x^{2} < \frac{225}{4} :

True for all

x

True for all x

Find singularity points

Find non-negative values for radicals:

- 1 \leq x \leq 5

Combine the intervals

True for all x and - 1 \leq x \leq 5

Merge Overlapping Intervals

- 1 \leq x \leq 5

\frac{3}{8} + \frac{1}{16}

\frac{4}{12}

3 = x^{2} - 2 x

\frac{11}{12} - \frac{5}{12}

\frac{x ^{2} - 1}{x ^{2} + 1} \leq 0

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