Square Root Problems: Problems with Solutions

Solution:
[tex]\sqrt{1} = 1[/tex]

Solution:
[tex]\sqrt{4} = \sqrt{2^2} = 2[/tex]

Solution:
[tex]\sqrt{25} = \sqrt{5^2} = 5[/tex]

Solution:
[tex]\sqrt{\frac{1}{4}}=\sqrt{\left(\frac{1}{2}\right)^2}=\frac{1}{2}[/tex]

Solution:
Convert the mixed number to an improper fraction: [tex]1\frac{24}{25}=\frac{49}{25}[/tex], [tex]\sqrt{\frac{49}{25}}=\frac{7}{5}=1. 4[/tex]
[tex]\sqrt{0. 09}=0. 3[/tex], [tex]\sqrt{3^2+4^2}=\sqrt{25}=5[/tex]
[tex]1. 4-0. 3+5=6. 1[/tex]
Be careful: [tex]\sqrt{3^2+4^2}\neq 3+4[/tex].

Solution:
The largest perfect-square factor of [tex]75[/tex] is [tex]25[/tex]:
[tex]\sqrt{75}=\sqrt{25}\,\sqrt{3}=5\sqrt{3}[/tex]

Solution:
The largest perfect-square factor of [tex]648[/tex] is [tex]324[/tex]:
[tex]\sqrt{648}=\sqrt{324}\,\sqrt{2}=18\sqrt{2}[/tex]

Solution:
Write [tex]0.2[/tex] as a fraction and multiply the numerator and the denominator by [tex]5[/tex]:
[tex]\sqrt{0.2}=\sqrt{\frac{1}{5}}=\sqrt{\frac{5}{25}}=\frac{\sqrt{5}}{5}[/tex]

Solution:
Multiply the numerator and the denominator under the root by [tex]2[/tex], so that the denominator becomes a perfect square:
[tex]\sqrt{\frac{3}{8}}=\sqrt{\frac{6}{16}}=\frac{\sqrt{6}}{4}[/tex]

Solution:
Multiply the numerator and the denominator under the root by [tex]10[/tex], so that the denominator becomes a perfect square:
[tex]\sqrt{\frac{3}{10}}=\sqrt{\frac{30}{100}}=\frac{\sqrt{30}}{10}[/tex]

Solution:
The largest perfect-square factor of [tex]200x^5[/tex] is [tex]100x^4[/tex] (here [tex]x\geq 0[/tex]):
[tex]\sqrt{200x^5}=\sqrt{100x^4}\,\sqrt{2x}=10x^2\sqrt{2x}[/tex]

Solution:
The largest perfect-square factor of [tex]12xz[/tex] is [tex]4[/tex]:
[tex]\sqrt{12xz}=\sqrt{4}\,\sqrt{3xz}=2\sqrt{3xz}[/tex]

Solution:
Multiply the numerator and the denominator by [tex]\sqrt{2}[/tex]:
[tex]\frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{\sqrt{2}\,\sqrt{2}}=\frac{3\sqrt{2}}{2}[/tex]

Solution:
Simplify the fraction first: [tex]\frac{5}{\sqrt{5}}=\frac{5\sqrt{5}}{5}=\sqrt{5}[/tex].
Therefore [tex]\left(\sqrt{5}+\sqrt{5}\right)^2=\left(2\sqrt{5}\right)^2=4\times 5=20[/tex].

Solution:
Take factors out from under the radical sign to get like radicals:
[tex]5\sqrt{12}=5\sqrt{4\times 3}=10\sqrt{3}[/tex], [tex]2\sqrt{27}=2\sqrt{9\times 3}=6\sqrt{3}[/tex], [tex]\sqrt{75}=\sqrt{25\times 3}=5\sqrt{3}[/tex]
[tex]10\sqrt{3}+6\sqrt{3}-5\sqrt{3}=11\sqrt{3}[/tex]

Solution:
The radicand of a square root must be non-negative:
[tex]9-3x\geq 0 \Rightarrow 3x\leq 9 \Rightarrow x\leq 3[/tex]
Therefore [tex]x\in(-\infty, 3][/tex].

Solution:
Bring the factors under the radical sign:
[tex]6\sqrt{2}=\sqrt{36\times 2}=\sqrt{72}[/tex], [tex]3\sqrt{7}=\sqrt{9\times 7}=\sqrt{63}[/tex], [tex]2\sqrt{14}=\sqrt{4\times 14}=\sqrt{56}[/tex]
[tex]\sqrt{56}\lt \sqrt{58}\lt \sqrt{63}\lt \sqrt{72}[/tex]
Therefore the smallest number is [tex]2\sqrt{14}[/tex].

Solution:
[tex]\sqrt{17^2-8^2}=\sqrt{(17-8)(17+8)}=\sqrt{9\times 25}=3\times 5=15[/tex]
[tex]\sqrt{(-2)^6}=\sqrt{64}=8[/tex]
[tex]\left(-\frac{2}{\sqrt{2}}\right)^2=\frac{4}{2}=2[/tex]
[tex]15-8-2=5[/tex]
Be careful: [tex]\sqrt{17^2-8^2}\neq 17-8[/tex].

Solution:
[tex]\sqrt{32}=4\sqrt{2}[/tex], [tex]\left(\sqrt{2}-1\right)^2=2-2\sqrt{2}+1=3-2\sqrt{2}[/tex]
[tex]4\sqrt{2}+3-2\sqrt{2}=2\sqrt{2}+3[/tex]
[tex]k=2[/tex], [tex]p=3[/tex], [tex]k+p=5[/tex]

Solution:
Bring the fractions to a common denominator; the denominator is a difference of squares:
[tex]\frac{\left(5-2\sqrt{6}\right)+\left(5+2\sqrt{6}\right)}{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}=\frac{10}{25-24}=10[/tex]

Solution:
Multiply the numerator and the denominator by the conjugate expression [tex]9-4\sqrt{5}[/tex]:
[tex]\frac{9-4\sqrt{5}}{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}=\frac{9-4\sqrt{5}}{81-16\times 5}=\frac{9-4\sqrt{5}}{1}=9-4\sqrt{5}[/tex]

Solution:
[tex]A\times B=\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)=2-1=1[/tex]
Since their product is [tex]1[/tex], the numbers are reciprocal. They are not equal, not opposite ([tex]A+B=2\sqrt{2}\neq 0[/tex]), and both are irrational.

Solution:
The mistake is in the step [tex](-1)+1=\sqrt{1}+\sqrt{1}[/tex]: we have [tex]\sqrt{1}=1[/tex], not [tex]-1[/tex].
By definition the square root of a number is non-negative, so [tex]-1\neq\sqrt{1}[/tex]. The correct statement is [tex](-1)+1=-\sqrt{1}+\sqrt{1}=0[/tex].

Solution:
Recall that [tex]\sqrt{a^2}=|a|[/tex].
[tex]4\sqrt{3}=\sqrt{48}\gt \sqrt{25}=5[/tex], so [tex]5-4\sqrt{3}\lt 0[/tex] and [tex]\sqrt{\left(5-4\sqrt{3}\right)^2}=4\sqrt{3}-5[/tex].
[tex]\left(\sqrt{3}-2\right)^2=3-4\sqrt{3}+4=7-4\sqrt{3}[/tex]
[tex]P=4\sqrt{3}-5+7-4\sqrt{3}=2[/tex]