Question

If $k=\sqrt{28+16\sqrt{3}}$, then what will be the value of $k^{1/4}-k^{-1/4}$?

• A. $\frac{2+\sqrt{3}}{2}$
• B. $\frac{2-\sqrt{3}}{2}$
• C. $\frac{3+\sqrt{3}}{2}$
• D. $\frac{-\sqrt{3}}{2}$
• E. $Noneofthese$

Answer 1

The correct option is C $\frac{3+\sqrt{3}}{2}$

Given:$k=\sqrt{28+16\sqrt{3}}$

Let $\sqrt{a}+\sqrt{b}=\sqrt{28+16\sqrt{3}}$

Squaring both sides, we get

$\Rightarrow a+b+2\sqrt{ab}=28+16\sqrt{3}$

Equating we get $a+b=28,2\sqrt{ab}=16\sqrt{3}$

$\Rightarrow a+b=28,\sqrt{ab}=8\sqrt{3}$

$\Rightarrow a+b=28,ab=64\times 3=192$

$\Rightarrow b=28-a,ab=192$

$\Rightarrow a(28-a)=192$

$\Rightarrow 28a-a^2-192=0$

$\Rightarrow a^2-28a+192=0$

$\Rightarrow a^2-16a-12a+192=0$

$\Rightarrow a(a-16)-12(a-16)=0$

$\Rightarrow (a-16)(a-12)=0$

$\Rightarrow a=16,12$

$\Rightarrow b=28-16,28-12$ since $b=28-a$

$\Rightarrow b=12,16$

$\therefore \sqrt{a}+\sqrt{b}=\sqrt{12}+\sqrt{16}=2\sqrt{3}+4$ for

$a=12,b=16$

and $\sqrt{a}+\sqrt{b}=\sqrt{16}+\sqrt{12}=4+2\sqrt{3}$ for $a=16,b=12$

$\therefore \sqrt{a}+\sqrt{b}=4+2\sqrt{3}$

Now ,$\sqrt{k}=\sqrt{a}+\sqrt{b}=4+2\sqrt{3}$

$k^{\frac{1}{4}}=\sqrt{\sqrt{a}+\sqrt{b}}=\sqrt{4+2\sqrt{3}}$

Let $k^{\frac{1}{4}}=\sqrt{c}+\sqrt{d}=\sqrt{4+2\sqrt{3}}$

Squaring both sides, we get

$\Rightarrow c+d+2\sqrt{cd}=4+2\sqrt{3}$

$\Rightarrow c+d=4,cd=3$

$\Rightarrow c=4-d,cd=3$

$\Rightarrow (4-d)d-3=0$

$\Rightarrow 4d-d^2-3=0$

$\Rightarrow -d^2+4d-3=0$

$\Rightarrow d^2-4d+3=0$

$\Rightarrow d^2-3d-d+3=0$

$\Rightarrow d(d-3)-1(d-3)$

$\Rightarrow d=3,1$

$\Rightarrow c=4-3,4-1$ since $c=4-d$

$\Rightarrow c=1,3$

$\therefore \sqrt{c}+\sqrt{d}=\sqrt{1}+\sqrt{3}=1+\sqrt{3}$ when $c=1,d=3$

and $\sqrt{c}+\sqrt{d}=\sqrt{3}+\sqrt{1}=\sqrt{3}+1$ when $c=3,d=1$

$\therefore k^{\frac{1}{4}}=1+\sqrt{3}$

Now, $k^{\frac{1}{4}}-k^{-\frac{1}{4}}$

$=k^{\frac{1}{4}}-\frac{1}{k^{\frac{1}{4}}}$

$=1+\sqrt{3}-\frac{1}{1+\sqrt{3}}$

$=\frac{1+3+2\sqrt{3}-1}{1+\sqrt{3}}$

$=\frac{3+2\sqrt{3}}{1+\sqrt{3}}$

$=\frac{3+2\sqrt{3}}{1+\sqrt{3}}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}$

$=\frac{3\sqrt{3}-3+6-2\sqrt{3}}{3-1}$

$=\frac{\sqrt{3}+3}{2}=\frac{3+\sqrt{3}}{2}$