What is 1+x>= 2sqrt(x) ?

The solution to 1+x>= 2sqrt(x) is x>= 0

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1+x>= 2sqrt(x)

1 + x \geq 2 x

x \geq 0

Interval Notation

[0, \infty)

Solution steps

1 + x \geq 2 x

Switch sides

2 x \leq 1 + x

Divide both sides by

2

\frac{2 x}{2} \leq \frac{1}{2} + \frac{x}{2}

Simplify

x \leq \frac{1 + x}{2}

f (x) \leq g (x)

then

g (x) \geq 0

\frac{1 + x}{2} \geq 0 : x \geq - 1

x \leq \frac{1 + x}{2} :

True for all

x

Find singularity points

Find non-negative values for radicals:

x \geq 0

Combine the intervals

x \geq - 1 and True for all x and x \geq 0

Merge Overlapping Intervals

x \geq 0

Verify Solutions:

x \geq 0

True

The solution is

x \geq 0

8^{2} - 2^{5} = x

0 = 205 - 11 t - 16 t^{2}

3 - 2 x - x^{2} = 0

7 x^{2} - 14 x - 21

\frac{- x}{4} - \frac{x}{3} = 14