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# equation

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

Jul 1, 2017
edited by Guest  Jul 1, 2017

#1
+2298
+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.

$$\frac{5\sqrt{3}}{3}\leq x\leq \frac{19\sqrt{3}}{3}$$

.
Jul 2, 2017

#1
+2298
+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