Respuesta :

TaeKwonDoIsDead

Answer:

[tex]x=4+-2\sqrt{7}[/tex]

Step-by-step explanation:

To solve, we need to take square root of both sides first:

[tex](x-4)^2=28\\\sqrt{(x-4)^2} =+-\sqrt{28} \\x-4=+-\sqrt{28}[/tex]

Now we use the property of radical shown below to simplify square root of 28:

Property:  [tex]\sqrt{x*y}=\sqrt{x} \sqrt{y}[/tex]

Now, we have:

[tex]x-4=+-\sqrt{28}\\x-4=+-\sqrt{4*7}\\x-4=+-\sqrt{4}\sqrt{7}\\x-4=+-2\sqrt{7}[/tex]

Now we take 4 to the other side and isolate x:

[tex]x-4=+-2\sqrt{7} \\x=4+-2\sqrt{7}[/tex]

first answer choice is right.

mixter17

Hello!

The answer is:

The correct answer is the first option.

[tex](x-4)^{2}=4+-2\sqrt{7}[/tex]

Why?

To solve the problem, we must remember the following square root property:

[tex]\sqrt{ab}=\sqrt{a}\sqrt{b}[/tex]

So,

We are given the expression:

[tex](x-4)^{2}=28[/tex]

Now, isolating we have:

First, applying square root to both sides of the equation in order to simplify the quadratic term, we have:

[tex]\sqrt{(x-4)^{2} }=\sqrt{28}\\(x-4)=\sqrt{4*7}\\\\x=4+-(\sqrt{4}*\sqrt{7})\\\\x=4+-2\sqrt{7}[/tex]

Hence, the correct answer is the first option.

[tex](x-4)^{2}=4+-2\sqrt{7}[/tex]

Have a nice day!

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