+0

# equation

0
59
1

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

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

#1
+815
+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
Sort:

#1
+815
+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

### 15 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details