Question

Solution of equation
√3X-2 = 2√3+4 is:​

Answer 1

Answer:

refer to the attachment..

Step-by-step explanation:

hope it helps you...

Answer 2

To solve: $\sqrt{3x}-2=2\sqrt{3}+4$

Solution:

Now, $\sqrt{3x}-2=2\sqrt{3}+4$

Adding 2 in both sides:

$\quad \sqrt{3x}-2+2=2\sqrt{3}+4+2$

$\to \sqrt{3x}=2\sqrt{3}+6$

Squaring both sides:

$\quad (\sqrt{3x})^{2}=(2\sqrt{3}+6)^{2}$

$\to 3x=12+24\sqrt{3}+36$

$\to 3x=48+24\sqrt{3}$

Dividing both sides by 3:

$\quad x=\frac{48+24\sqrt{3}}{3}$

$\to x=16+8\sqrt{3}$

Therefore the required solution is

$\quad\quad\quad x=16+8\sqrt{3}$

Another assumption:

To solve: $\sqrt{3x}-2=2\sqrt{3}+4$

Solution:

Now, $\sqrt{3x-2}=2\sqrt{3}+4$

Squaring both sides, we get:

$\quad 3x-2=(2\sqrt{3}+4)^{2}$

$\to 3x-2=12+16\sqrt{3}+16$

Taking x terms in LHS and constants in RHS:

$\quad 3x=12+16\sqrt{3}+16+2$

$\to 3x=30+16\sqrt{3}$

Divide both sides by 3:

$\quad x=\frac{30+16\sqrt{3}}{3}$

Therefore the required solution is

$\quad\quad\quad x=\frac{30+16\sqrt{3}}{3}$

Note: To solve this type of problems, remember to investigate if the problem can be solved just by squaring and rearranging the terms, or not.

Related problem:

Q. Find the positive root of √3x²+6=9

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