What is 7sqrt(x)+1<9 ?

The solution to 7sqrt(x)+1<9 is 0<= x< 64/49

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7sqrt(x)+1<9

7 x + 1 < 9

0 \leq x < \frac{64}{49}

Interval Notation

[0, \frac{64}{49})

Decimal Notation

0 \leq x < 1.30612 \dots

Solution steps

7 x + 1 < 9

Move

1

to the right side

7 x < 8

Divide both sides by

7

x < \frac{8}{7}

Square both sides

(x)^{2} < (\frac{8}{7})^{2}

Simplify

x < (\frac{8}{7})^{2}

Expand

(\frac{8}{7})^{2} : \frac{64}{49}

x < \frac{64}{49}

Find singularity points

Find non-negative values for radicals:

x \geq 0

Combine the intervals

x < \frac{64}{49} and x \geq 0

Merge Overlapping Intervals

0 \leq x < \frac{64}{49}

\frac{3}{1 + i} - \frac{4}{1 - i}

(4 x^{- 2} y^{4})^{- \frac{1}{2}}

297

\frac{25 !}{7 ! ( 25 - 7 ) !}