Square Root Problems: Very Difficult Problems with Solutions

Solution:
Rationalize each term:
[tex]\frac{1}{\sqrt{k}+\sqrt{k+1}}=\frac{\sqrt{k+1}-\sqrt{k}}{(k+1)-k}=\sqrt{k+1}-\sqrt{k}[/tex]
All intermediate terms cancel:
[tex]\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+\ldots+\left(\sqrt{100}-\sqrt{99}\right)=\sqrt{100}-\sqrt{1}=10-1=9[/tex]

Solution:
[tex]\sqrt{24}=2\sqrt{6}[/tex], [tex]\sqrt{40}=2\sqrt{10}[/tex], [tex]\sqrt{60}=2\sqrt{15}[/tex]
Compare with [tex]\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\sqrt{ab}+2\sqrt{ac}+2\sqrt{bc}[/tex].
It fits for [tex]a=2[/tex], [tex]b=3[/tex], [tex]c=5[/tex]: [tex]a+b+c=10[/tex], [tex]ab=6[/tex], [tex]ac=10[/tex], [tex]bc=15[/tex].
[tex]\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{2}+\sqrt{3}+\sqrt{5}[/tex]

Solution:
Let [tex]t=\sqrt{x-1}\geq 0[/tex], then [tex]x=t^2+1[/tex]:
[tex]x\pm 2\sqrt{x-1}=t^2\pm 2t+1=(t\pm 1)^2[/tex]
[tex]\sqrt{(t-1)^2}+\sqrt{(t+1)^2}=|t-1|+t+1[/tex]
For [tex]1\leq x\leq 2[/tex] we have [tex]0\leq t\leq 1[/tex], so [tex]|t-1|=1-t[/tex]:
[tex]1-t+t+1=2[/tex]

Solution:
Work from the innermost radical outward:
[tex]9+4\sqrt{2}=1+4\sqrt{2}+8=\left(1+2\sqrt{2}\right)^2[/tex], so [tex]\sqrt{9+4\sqrt{2}}=1+2\sqrt{2}[/tex].
[tex]2+1+2\sqrt{2}=3+2\sqrt{2}=\left(1+\sqrt{2}\right)^2[/tex], so [tex]\sqrt{3+2\sqrt{2}}=1+\sqrt{2}[/tex].
[tex]13+30\left(1+\sqrt{2}\right)=43+30\sqrt{2}=25+30\sqrt{2}+18=\left(5+3\sqrt{2}\right)^2[/tex]
[tex]5+3\sqrt{2}-\sqrt{18}=5+3\sqrt{2}-3\sqrt{2}=5[/tex]

Solution:
[tex]x^2\mp\sqrt{20}\,x+5=x^2\mp 2\sqrt{5}\,x+5=\left(x\mp\sqrt{5}\right)^2[/tex], so the expression equals [tex]\left|x-\sqrt{5}\right|-\left|x+\sqrt{5}\right|[/tex].
[tex]\sqrt{2+\sqrt{2+\sqrt{3}}}\times \sqrt{2-\sqrt{2+\sqrt{3}}}=\sqrt{4-\left(2+\sqrt{3}\right)}=\sqrt{2-\sqrt{3}}[/tex]
[tex]x=\sqrt{2+\sqrt{3}}\times \sqrt{2-\sqrt{3}}=\sqrt{4-3}=1[/tex]
[tex]\left|1-\sqrt{5}\right|-\left|1+\sqrt{5}\right|=\sqrt{5}-1-1-\sqrt{5}=-2[/tex]

Solution:
[tex]2\pm\sqrt{3}=\frac{4\pm 2\sqrt{3}}{2}=\frac{\left(\sqrt{3}\pm 1\right)^2}{2}[/tex], so [tex]\sqrt{2\pm\sqrt{3}}=\frac{\sqrt{3}\pm 1}{\sqrt{2}}[/tex].
[tex]\sqrt{2}+\frac{\sqrt{3}+1}{\sqrt{2}}=\frac{3+\sqrt{3}}{\sqrt{2}}[/tex], [tex]\sqrt{2}-\frac{\sqrt{3}-1}{\sqrt{2}}=\frac{3-\sqrt{3}}{\sqrt{2}}[/tex]
[tex]\sqrt{2}\left(\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\right)=\sqrt{2}\times \frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{9-3}[/tex]
[tex]=\sqrt{2}\times \frac{\left(3+\sqrt{3}\right)+\left(3-\sqrt{3}\right)}{6}=\sqrt{2}[/tex]

Solution:
Let [tex]u=\sqrt{a}+\sqrt{a-1}[/tex]. Then [tex]\frac{1}{u}=\sqrt{a}-\sqrt{a-1}[/tex], because [tex]\left(\sqrt{a}+\sqrt{a-1}\right)\left(\sqrt{a}-\sqrt{a-1}\right)=a-(a-1)=1[/tex].
[tex]u-\frac{1}{u}=2\sqrt{a-1}[/tex], [tex]u+\frac{1}{u}=2\sqrt{a}[/tex]
[tex]u^4-1=u^2\left(u^2-\frac{1}{u^2}\right)=u^2\left(u-\frac{1}{u}\right)\left(u+\frac{1}{u}\right)=4u^2\sqrt{a}\sqrt{a-1}[/tex]
[tex]\frac{4\sqrt{a}\,u^2}{4u^2\sqrt{a}\sqrt{a-1}}=\frac{1}{\sqrt{a-1}}[/tex]

Solution:
Denote the numerator by [tex]A[/tex] and the denominator by [tex]B[/tex]. We use the identity (for [tex]1\leq k\leq 99[/tex]):
[tex]\sqrt{10\pm\sqrt{k}}=\sqrt{\frac{10+\sqrt{100-k}}{2}}\pm\sqrt{\frac{10-\sqrt{100-k}}{2}}[/tex]
It can be checked by squaring the right side: [tex]\frac{10+\sqrt{100-k}}{2}+\frac{10-\sqrt{100-k}}{2}\pm 2\sqrt{\frac{100-(100-k)}{4}}=10\pm\sqrt{k}[/tex].
When [tex]k[/tex] runs from [tex]1[/tex] to [tex]99[/tex], so does [tex]100-k[/tex]. Hence
[tex]A=\frac{1}{\sqrt{2}}\sum_{k=1}^{99}\left(\sqrt{10+\sqrt{k}}+\sqrt{10-\sqrt{k}}\right)=\frac{A+B}{\sqrt{2}}[/tex]
[tex]A\sqrt{2}=A+B \Rightarrow B=A\left(\sqrt{2}-1\right)[/tex]
[tex]\frac{A}{B}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1[/tex]