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2 years ago

PLEASE SOLVE AND CHECK. SHOW COMPLETE SOLUTION

Mathematics

2 answers:

klasskru [66]2 years ago

5 0

Solution:

$\sqrt{x + 4} \geqslant 2 \sqrt{x}$

Square both sides.

$= > ( \sqrt{x + 4)} ^{2} = (2 \sqrt{x} ) ^{2} \\$

In the LHS, square and square root gets cancelled out. In the RHS, do the square.

$= > x + 4 \geqslant {2}^{2} x \\ = > x + 4 \geqslant 4x$

Transpose x to the RHS.

$= > 4 \geqslant 4x - x \\ = > 3x \leqslant 4$

Now, divide both sides by 3.

$= > \frac{3x}{3} \leqslant \frac{4}{3} \\ = > x \leqslant \frac{4}{3}$

Answer:

$x \leqslant \frac{4}{3}$

Hope you could understand.

If you have any query, feel free to ask.

Paul [167]2 years ago

3 0

Answer:

4/3 $\geq$ x

Step-by-step explanation:

$(\sqrt{x+4})^ {2} \geq (2\sqrt{x})^{2}\\$

x + 4 $\geq$ 4x

4 $\geq$ 4x - x

4 $\geq$ 3x

4/3 $\geq$ x

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This was not copied from a website or someone else. This was from my last year report.


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