Solving a Radical Equation

We have $\left(3+5\sqrt{3}\right)^2=84+30\sqrt{3}$

and $\left(9-\sqrt{3}\right)^2 = 84-18\sqrt{3}$ .

So $\begin{aligned} N & = \sqrt{84+30\sqrt{3}}+5\sqrt{84-18\sqrt{3}}\\ & = |3+5\sqrt{3}|+5|9-\sqrt{3}|=3+5\sqrt{3}+45-5\sqrt{3}=48 \end{aligned}$

We split up the problem into two parts. $\sqrt{84 + 30 \sqrt{3}}$ and $5( \sqrt{84 - 18 \sqrt{3}}$ .

We solve $5(\sqrt{84 - 18 \sqrt{3}})$ by molding it into a perfect square. Notice we can break down the expression $\sqrt{84 - 18 \sqrt{3}}$ into $\sqrt{ 81 + 3 - 2(9)(\sqrt{3}) }$ . This turns out to be $(9 - \sqrt{3}) ^2$ . Putting it back into the original expression, we get $5(\sqrt{(9 - \sqrt{3})^2})$ which is $45 - 5 \sqrt{3}$ .

We solve the second part of the equation the same way. We know to work have to work with $5 \sqrt{3}$ to get an integer so we break down the expression into $\sqrt{(75 + 9 + 2 \times 3 \times (5 \sqrt{3})}$ . This turns into $\sqrt{(3 + 5 \sqrt{3})^2}$ which is $3 + 5 \sqrt{3}$ .

Add these two final expressions together and we get $45 - 5 \sqrt{3}$ + $3 + 5 \sqrt{3}$ . This cleans up nicely to equal $48$ .

$N =\sqrt{84+30\sqrt{3}} + 5\sqrt{84-18\sqrt{3}}=$ $\sqrt{3^2 + 2.3.5\sqrt{3} + (5\sqrt{3})^2} + 5\sqrt{9^2-2.9.\sqrt{3} + (\sqrt{3})^2}=$ $\sqrt{(3+5\sqrt{3})^2}+5\sqrt{(9-\sqrt{3})^2} =3+5\sqrt{3} + 5(9-\sqrt{3})=48$ .

\sqrt{84+30 \cdot \sqrt{3}} can be written as: \sqrt{9+75+30 \cdot \sqrt{3}} \sqrt{{5 \cdot \sqrt{3}}^2+30 \cdot \sqrt{3} +3^2} which is a perfect square trinomial that can be factored as \sqrt{{5 \cdot \sqrt {3} +3}^2} then cancel the radical 5 \cdot \sqrt {3} +3

the same as the other term: it can be simplify as 5 \cdot \sqrt{{9 -\sqrt{3}}^2} then 5 \cdot {9 -\sqrt{3}} then add the two terms: the 5 \cdot \sqrt {3} will be canceled so the answer is 45 +3 = 48

Squaring both side, we have N^2 = 84+30 rt3+25 84-25 18 rt3 + 10 rt(84 84 -30 18 3+84 (30-18) rt3) N^2 = 26 84-420 rt3 + 10 rt(5436-1008 rt3) N^2 = 26 84-420 rt3 + 10 (12+42 rt3) N^2 = 2304 N=48

We square both sides to obtain N^{2} = 84 + 30\sqrt{3} + 25(84-18\sqrt{3}) + 10\sqrt{84^{2} + 30 * 84 \sqrt{3} - 18 * 84\sqrt{3} - 18 * 3 * 30}. We can simplify this to N^{2} - 2184 + 420\sqrt{3} = 60\sqrt{151 + 28\sqrt{3}}.

Squaring this result again, we obtain 5299056-1834560 \sqrt{3}-4368 N^{2}+840 \sqrt(3) N^{2}+N^{4} = 60^2(151+28\sqrt{3}).

Solving this quartic, we obtain (N-48)(N+48)(N^{2}+840\sqrt{3} - 2064). N is positive, thus N = 48.

Solution 1: We can complete the square for both the terms: $84 + 30\sqrt{3} = 9 + 30\sqrt{3} + 75 = (3 + 5\sqrt{3})^2$ and $84 - 18\sqrt{3} = 81 - 18\sqrt{3} + 3 = (9 - \sqrt{3})^2$ . Since square roots are positive, we have $\begin{aligned} N &= \sqrt{84 + 30\sqrt{3} } + 5 \sqrt{ 84 - 18 \sqrt{3} } \\ &= \sqrt { (3+ 5 \sqrt{3})^2 } + 5 \sqrt{ (9 - \sqrt{3} )^2 } \\ &= (3 + 5 \sqrt{3} ) + 5 ( 9 -\sqrt{3} ) \\ &= 48 \\ \end{aligned}$

Solution 2: We guess that $84+30\sqrt{3} = (a + b \sqrt{3})^2 = a^2 + 3b^2 + 2 ab \sqrt{3}$ , which gives $84 = a^2 + 3b^2$ and $30 = 2ab$ . Substituting the second equation into the first gives $a^2 + 3 (\frac {15}{a})^2 = 84$ $\Rightarrow a^4 - 84 a^2 + 3 \cdot 225 = 0$ $\Rightarrow (a^2-9)(a^2-75)=0$ . This has solutions $a = \pm 3, \pm 5\sqrt{3}$ . We choose $a=3$ and get $b= \frac {30}{2\times 3}= 5$ . Hence, $84+30\sqrt{3} = (3 + 5 \sqrt{3})^2$

Similarly, we guess that $84 - 18\sqrt{3} =(c-d\sqrt{3})^2 = c^2 + 3d^2 - 2cd \sqrt{3}$ , which gives $84=c^2 +3d^2$ and $18=2cd$ . Substituting the second equation into the first gives $c^2 + 3(\frac {9}{c})^2=84$ $\Rightarrow c^4 - 84 c^2 + 3\cdot 81 =0$ $\Rightarrow (c^2-81)(c^2-3)=0$ . This has solutions $c=\pm 9, \pm \sqrt{3}$ . We choose $c=9$ and get $d = \frac {18}{2\times 9} = 1$ . hence $84 - 18\sqrt{3} = (9 - \sqrt{3})^2$ .

Since square roots are positive, we have $\begin{aligned} N &= \sqrt{84 + 30\sqrt{3} } + 5 \sqrt{ 84 - 18 \sqrt{3} } \\ &= \sqrt { (3+ 5 \sqrt{3})^2 } + 5 \sqrt{ (9 - \sqrt{3} )^2} \\ &= (3 + 5 \sqrt{3} ) + 5 ( 9 -\sqrt{3} ) \\ &= 48 \\ \end{aligned}$

Assume $\color{#D61F06}{84+30\sqrt{3}\;and\;84-18\sqrt{3}}$ to be perfect squares.First solve for $\color{#3D99F6}{84+30\sqrt{3}}$ : $\color{#20A900}{(\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{ab}=84+30\sqrt{3}\\ \begin{cases} a+b=84\\2\sqrt{ab}=30\sqrt{3}\rightarrow ab=675\end{cases}}$ Solving for $a,b$ gives us $\color{#624F41}{a=3,b=5\sqrt{3}}$ .So $\color{burntorange}{84+30\sqrt{3}=(3+5\sqrt3)^2}$ .So the first radical simplifies to $\color{#69047E}{3+5\sqrt3}$ .Solve for $\color{midnightblue}{84-18\sqrt3}$ : $\color{redviolet}{(\sqrt{a}-\sqrt{b})^2=a+b-2\sqrt{ab}=84-18\sqrt{3}\\ \begin{cases} a+b=84\\ -2\sqrt{ab}=-18\sqrt{3}\rightarrow ab=243\end{cases}}$ Solving,we get $\color{#EC7300}{a=3,b=9}$ .So second radical is $\color{#D61F06}{\sqrt{(\sqrt{3}-9)^2}}$ .As using these values of $a,b$ will not give an integral solution,so the second radical is $\color{#D61F06}{\sqrt{(9-\sqrt{3})^2}}$ because $\color{#69047E}{(x-y)^2=(y-x)^2}$ .So the question simplifies to $\color{burntorange}{3+5\sqrt{3}+5(9-\sqrt{3})=3+45+5\sqrt{3}-5\sqrt{3}=\boxed{48}}$

sqrt 3 is approx 1.7 when multiplied to 30, approx 45 18, approx 27

84+45= 129. sqrt of this is > 12 84-27= 57. 5 sqrt is > 7 5

the sum should be greater that 35+12= 47 so, 48

48