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(x-sqroot(48))^2<=49/3 ??

Guest Jul 1, 2017
edited by Guest  Jul 1, 2017

Best Answer 

 #1
avatar+563 
+2

\((x-\sqrt{48})^2\leq\frac{49}{3}\) 

 

I am sorry if I have interpreted your inequality incorrectly. I am fairly certain that the equation above is what you were attempting to convey.

 

\((x-\sqrt{48})^2\leq\frac{49}{3}\) Undo the square by square rooting both sides of the inequality.
\(x-\sqrt{48}\leq|\sqrt{\frac{49}{3}}|\) Remember the square root rule that states that \(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\). I like to think of it as "distributing" the square root into both the numerator and denominator. Also, the square root function always results in a negative and positive answer. That's why the absolute value sign is there.
\(x-\sqrt{48}\leq|\frac{\sqrt{49}}{\sqrt{3}}|\) Simplify the fraction on the right side of the equation.
\(x-\sqrt{48}\leq|\frac{7}{\sqrt{3}}|\) Let's rationalize the right hand side of the equation by multiplying the numerator and denominator by 
\(\frac{7}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}=\frac{7\sqrt{3}}{3}\) Reinsert this back into the inequality.
\(x-\sqrt{48}\leq|\frac{7\sqrt{3}}{3}|\) The absolute value results in 2 answers: the positive and negative. Let's split that up.
\(x-\sqrt{48}\leq\pm\frac{7\sqrt{3}}{3}\)  

 

Now, we must evaluate each inequality separately. The one for the plus and the one for the minus. To make it easier, I'll solve both one at a time! I'll do the positive version first:

 

\(x-\sqrt{48}\leq\frac{7\sqrt{3}}{3}\) Add \(\sqrt{48}\) to both sides.
\(x\leq\frac{7\sqrt{3}}{3}+\sqrt{48}\) First, I'll make\(\sqrt{48}\) into simplest radical form.
\(\sqrt{48}=\sqrt{16*3}=\sqrt{16}\sqrt{3}=4\sqrt{3}\) Ok, because this radical is in simplest radical form, reinsert it back into the equation. 
\(x\leq\frac{7\sqrt{3}}{3}+\frac{4\sqrt{3}}{1}\) Put \(\frac{4\sqrt{3}}{1}\) into a common denominator so that it is possible to add them together.
\(\frac{4\sqrt{3}}{1}*\frac{3}{3}=\frac{12\sqrt{3}}{3}\) Now that this fraction has a common denominator, let's add them together.
\(x\leq\frac{7\sqrt{3}}{3}+\frac{12\sqrt{3}}{3}\) Add the fractions together. 
\(x\leq\frac{19\sqrt{3}}{3}\) Time to solve the second part of the equation.
   

 

Now, we have to solve for the other scenario, \(x-\sqrt{48}\geq-\frac{7\sqrt{3}}{3}\). Now, this is a tad difficult to explain, but the absolute value function does funny things to an inequality. The rule is that \(|f(x)|\leq a\Rightarrow f(x)\leq a\hspace{1mm}\text{and}\hspace{1mm}f(x)\geq -a\)

 

\(x-\sqrt{48}\geq-\frac{7\sqrt{3}}{3}\) Solve and isolate x by adding \(\sqrt{48}\) to both sides.
\(x\geq-\frac{7\sqrt{3}}{3}+\sqrt{48}\) Put \(\sqrt{48}\) into simplest radical form and put in common denomator. Because I have already done this in the previous equation, I'll just skip.
\(x\geq-\frac{7\sqrt{3}}{3}+\frac{12\sqrt{3}}{3}\) Add the fractions together now.
\(x\geq\frac{5\sqrt{3}}{3}\)  
   

 

Now, put the solutions together.

 

\(\frac{5\sqrt{3}}{3}\leq x\leq \frac{19\sqrt{3}}{3}\)

TheXSquaredFactor  Jul 2, 2017
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1+0 Answers

 #1
avatar+563 
+2
Best Answer

\((x-\sqrt{48})^2\leq\frac{49}{3}\) 

 

I am sorry if I have interpreted your inequality incorrectly. I am fairly certain that the equation above is what you were attempting to convey.

 

\((x-\sqrt{48})^2\leq\frac{49}{3}\) Undo the square by square rooting both sides of the inequality.
\(x-\sqrt{48}\leq|\sqrt{\frac{49}{3}}|\) Remember the square root rule that states that \(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\). I like to think of it as "distributing" the square root into both the numerator and denominator. Also, the square root function always results in a negative and positive answer. That's why the absolute value sign is there.
\(x-\sqrt{48}\leq|\frac{\sqrt{49}}{\sqrt{3}}|\) Simplify the fraction on the right side of the equation.
\(x-\sqrt{48}\leq|\frac{7}{\sqrt{3}}|\) Let's rationalize the right hand side of the equation by multiplying the numerator and denominator by 
\(\frac{7}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}=\frac{7\sqrt{3}}{3}\) Reinsert this back into the inequality.
\(x-\sqrt{48}\leq|\frac{7\sqrt{3}}{3}|\) The absolute value results in 2 answers: the positive and negative. Let's split that up.
\(x-\sqrt{48}\leq\pm\frac{7\sqrt{3}}{3}\)  

 

Now, we must evaluate each inequality separately. The one for the plus and the one for the minus. To make it easier, I'll solve both one at a time! I'll do the positive version first:

 

\(x-\sqrt{48}\leq\frac{7\sqrt{3}}{3}\) Add \(\sqrt{48}\) to both sides.
\(x\leq\frac{7\sqrt{3}}{3}+\sqrt{48}\) First, I'll make\(\sqrt{48}\) into simplest radical form.
\(\sqrt{48}=\sqrt{16*3}=\sqrt{16}\sqrt{3}=4\sqrt{3}\) Ok, because this radical is in simplest radical form, reinsert it back into the equation. 
\(x\leq\frac{7\sqrt{3}}{3}+\frac{4\sqrt{3}}{1}\) Put \(\frac{4\sqrt{3}}{1}\) into a common denominator so that it is possible to add them together.
\(\frac{4\sqrt{3}}{1}*\frac{3}{3}=\frac{12\sqrt{3}}{3}\) Now that this fraction has a common denominator, let's add them together.
\(x\leq\frac{7\sqrt{3}}{3}+\frac{12\sqrt{3}}{3}\) Add the fractions together. 
\(x\leq\frac{19\sqrt{3}}{3}\) Time to solve the second part of the equation.
   

 

Now, we have to solve for the other scenario, \(x-\sqrt{48}\geq-\frac{7\sqrt{3}}{3}\). Now, this is a tad difficult to explain, but the absolute value function does funny things to an inequality. The rule is that \(|f(x)|\leq a\Rightarrow f(x)\leq a\hspace{1mm}\text{and}\hspace{1mm}f(x)\geq -a\)

 

\(x-\sqrt{48}\geq-\frac{7\sqrt{3}}{3}\) Solve and isolate x by adding \(\sqrt{48}\) to both sides.
\(x\geq-\frac{7\sqrt{3}}{3}+\sqrt{48}\) Put \(\sqrt{48}\) into simplest radical form and put in common denomator. Because I have already done this in the previous equation, I'll just skip.
\(x\geq-\frac{7\sqrt{3}}{3}+\frac{12\sqrt{3}}{3}\) Add the fractions together now.
\(x\geq\frac{5\sqrt{3}}{3}\)  
   

 

Now, put the solutions together.

 

\(\frac{5\sqrt{3}}{3}\leq x\leq \frac{19\sqrt{3}}{3}\)

TheXSquaredFactor  Jul 2, 2017

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