Square Root Problems: Difficult Problems with Solutions

Solution:
The numbers [tex]2000[/tex] and [tex]2014[/tex] are symmetric about [tex]2007[/tex]:
[tex]2000\times 2014=(2007-7)(2007+7)=2007^2-49[/tex]
[tex]\sqrt{2007^2-49+49}=\sqrt{2007^2}=2007[/tex]

Solution:
[tex]\sqrt{8x^3y^2}=\sqrt{4x^2y^2\times 2x}=2|x||y|\sqrt{2x}[/tex]
Since [tex]x\gt 0[/tex], [tex]|x|=x[/tex], and since [tex]y\lt 0[/tex], [tex]|y|=-y[/tex]. Therefore [tex]\sqrt{8x^3y^2}=2x(-y)\sqrt{2x}=-2xy\sqrt{2x}[/tex].

Solution:
Write the radicands as perfect squares:
[tex]29\pm 12\sqrt{5}=9\pm 2\times 3\times 2\sqrt{5}+20=\left(3\pm 2\sqrt{5}\right)^2[/tex]
[tex]\sqrt{29-12\sqrt{5}}=\left|3-2\sqrt{5}\right|=2\sqrt{5}-3[/tex], because [tex]2\sqrt{5}=\sqrt{20}\gt 3[/tex].
[tex]\sqrt{29+12\sqrt{5}}=3+2\sqrt{5}[/tex]
[tex]\left(2\sqrt{5}-3\right)-\left(3+2\sqrt{5}\right)=-6[/tex]

Solution:
Let [tex]A=\sqrt{7+\sqrt{13}}+\sqrt{7-\sqrt{13}}\gt 0[/tex]. Square it:
[tex]A^2=\left(7+\sqrt{13}\right)+2\sqrt{\left(7+\sqrt{13}\right)\left(7-\sqrt{13}\right)}+\left(7-\sqrt{13}\right)[/tex]
[tex]A^2=14+2\sqrt{49-13}=14+2\times 6=26[/tex]
Therefore [tex]A=\sqrt{26}[/tex].

Solution:
[tex]\sqrt{38}+\sqrt{19}=\sqrt{19}\left(\sqrt{2}+1\right)[/tex]
[tex]m=\sqrt{19}\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)=\sqrt{19}(2-1)=\sqrt{19}[/tex]
[tex]n=3\sqrt{2}=\sqrt{18}[/tex]
[tex]\sqrt{19}\gt \sqrt{18}[/tex], so [tex]m\gt n[/tex].

Solution:
Group the denominator as [tex]\left(1+\sqrt{5}\right)+\sqrt{6}[/tex] and multiply the numerator and the denominator by [tex]\left(1+\sqrt{5}\right)-\sqrt{6}[/tex]:
[tex]\left(1+\sqrt{5}\right)^2-6=1+2\sqrt{5}+5-6=2\sqrt{5}[/tex]
[tex]\frac{10\left(1+\sqrt{5}-\sqrt{6}\right)}{2\sqrt{5}}=\frac{5\left(1+\sqrt{5}-\sqrt{6}\right)}{\sqrt{5}}=\sqrt{5}\left(1+\sqrt{5}-\sqrt{6}\right)=5+\sqrt{5}-\sqrt{30}[/tex]

Solution:
[tex]\left(\sqrt{10}-\sqrt{6}\right)^2=10-2\sqrt{60}+6=16-4\sqrt{15}=4\left(4-\sqrt{15}\right)[/tex]
Since [tex]\sqrt{10}-\sqrt{6}\gt 0[/tex], so [tex]\sqrt{4-\sqrt{15}}=\frac{\sqrt{10}-\sqrt{6}}{2}[/tex].
[tex]\left(4+\sqrt{15}\right)\frac{\left(\sqrt{10}-\sqrt{6}\right)^2}{2}=\left(4+\sqrt{15}\right)\frac{4\left(4-\sqrt{15}\right)}{2}=2\left(16-15\right)=2[/tex]

Solution:
Multiply the numerator and the denominator by [tex]\sqrt{(1+a)(1-a)}\gt 0[/tex]. Since [tex]\sqrt{\frac{1+a}{1-a}}\times \sqrt{(1+a)(1-a)}=1+a[/tex] and [tex]\sqrt{\frac{1-a}{1+a}}\times \sqrt{(1+a)(1-a)}=1-a[/tex], we get
[tex]\frac{(1+a)-(1-a)}{(1+a)+(1-a)}=\frac{2a}{2}=a[/tex]

Solution:
[tex]1+\left(\frac{a^2-1}{2a}\right)^2=\frac{4a^2+a^4-2a^2+1}{4a^2}=\frac{\left(a^2+1\right)^2}{4a^2}[/tex]
[tex]\sqrt{\frac{\left(a^2+1\right)^2}{4a^2}}=\frac{a^2+1}{2|a|}=\frac{a^2+1}{-2a}[/tex], because [tex]a\lt 0[/tex].
[tex]M=\frac{a}{a^2+1}\times \frac{a^2+1}{-2a}=-\frac{1}{2}[/tex]
A common mistake is to write [tex]\sqrt{4a^2}=2a[/tex] and get [tex]\frac{1}{2}[/tex]; for [tex]a\lt 0[/tex] we have [tex]\sqrt{4a^2}=2|a|=-2a[/tex].

Solution:
Assume the opposite: [tex]\sqrt{5}=\frac{p}{q}[/tex], where [tex]p[/tex] and [tex]q[/tex] are natural numbers and the fraction is irreducible. Then [tex]5=\frac{p^2}{q^2} \Rightarrow p^2=5q^2[/tex].
So [tex]p^2[/tex] is divisible by [tex]5[/tex]. Since [tex]5[/tex] is a prime number, [tex]p[/tex] is divisible by [tex]5[/tex]: [tex]p=5k[/tex]. Then [tex]25k^2=5q^2 \Rightarrow q^2=5k^2[/tex].
So [tex]q[/tex] is also divisible by [tex]5[/tex], and the fraction [tex]\frac{p}{q}[/tex] can be reduced by [tex]5[/tex] — a contradiction. Therefore [tex]\sqrt{5}[/tex] is irrational.

Solution:
Let [tex]x+\sqrt{2}=q[/tex], where [tex]q[/tex] is rational. Then [tex]x=q-\sqrt{2}[/tex] and [tex]x\sqrt{2}=\left(q-\sqrt{2}\right)\sqrt{2}=q\sqrt{2}-2[/tex].
This number is rational only if [tex]q\sqrt{2}[/tex] is rational. If [tex]q\neq 0[/tex], then [tex]\sqrt{2}=\frac{q\sqrt{2}}{q}[/tex] would be rational, which is false. Hence [tex]q=0[/tex].
So the only such number is [tex]x=-\sqrt{2}[/tex]. Check: [tex]x\sqrt{2}=-2[/tex] and [tex]x+\sqrt{2}=0[/tex].