What is sqrt(x)+5x<= 4 ?

The solution to sqrt(x)+5x<= 4 is 0<= x<= 16/25

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sqrt(x)+5x<= 4

x + 5 x \leq 4

0 \leq x \leq \frac{16}{25}

Interval Notation

[0, \frac{16}{25}]

Decimal Notation

0 \leq x \leq 0.64

Solution steps

x + 5 x \leq 4

Subtract

5 x

from both sides

x + 5 x - 5 x \leq 4 - 5 x

Simplify

x \leq 4 - 5 x

f (x) \leq g (x)

then

g (x) \geq 0

4 - 5 x \geq 0 : x \leq \frac{4}{5}

x \leq 4 - 5 x : x \leq \frac{16}{25} or x \geq 1

Find singularity points

Find non-negative values for radicals:

x \geq 0

Combine the intervals

x \leq \frac{4}{5} and (x \leq \frac{16}{25} or x \geq 1) and x \geq 0

Merge Overlapping Intervals

0 \leq x \leq \frac{16}{25}

Verify Solutions:

0 \leq x \leq \frac{16}{25}

True

The solution is

0 \leq x \leq \frac{16}{25}

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t^{2} - 16 = 0

- 3 y + 12 = - 48

x + 8 \geq 6