Exercises: Computing with Square Roots

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Simplify as far as possible.

$\left(1-\sqrt3\right)\cdot\left(1+\sqrt3\right)$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers

$\displaystyle \left(1-\sqrt{3}\right)\cdot\left(1+\sqrt{3}\right)$	$=$
	↓	Use the binomial formulas.
	$=$	$\displaystyle \left(1^2-\sqrt{3}^2\right)$
	↓	Square and root cancel.
	$=$	$\displaystyle \left(1-3\right)$
	$=$	$\displaystyle -2$

Do you have a question?

$\left(\sqrt2-3\sqrt2\right)^2$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers

$\displaystyle \left(\sqrt{2}-3\sqrt{2}\right)^2$	$=$
	↓	Subtract inside the bracket
	$=$	$\displaystyle \left(-2\sqrt{2}\right)^2$
	$=$	$\displaystyle \left(-2\right)^2\cdot\left(\sqrt{2}\right)^2$
	↓	Square both parts.
	$=$	$\displaystyle 4\cdot2$
	$=$	$\displaystyle 8$

$\sqrt3\cdot\left(\frac16\sqrt{12}-3\sqrt{\frac1{27}}\right)$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers

$\displaystyle \sqrt{3}\cdot\left(\frac{1}{6}\sqrt{12}-3\sqrt{\frac{1}{27}}\right)$	$=$
	↓	Pull $\frac{1}{6}$ and $3$ into the root.
	$=$	$\displaystyle \sqrt{3}\cdot\left(\sqrt{\frac{1}{36}\cdot12}-\sqrt{9\cdot\frac{1}{27}}\right)$
	↓	Multiply inside the roots.
	$=$	$\displaystyle \sqrt{3}\cdot\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{1}{3}}\right)$
	↓	Subtract inside the bracket.
	$=$	$\displaystyle \sqrt{3}\cdot0$
	$=$	$\displaystyle 0$

$\left(2\sqrt{108}-7\sqrt{54}\right):\sqrt{27}$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers

$\displaystyle \left(2\sqrt{108}-7\sqrt{54}\right):\sqrt{27}$	$=$
	↓	Convert the ":" into a fraction.
	$=$	$\displaystyle \frac{\left(2\sqrt{108}-7\sqrt{54}\right)}{\sqrt{27}}$
	↓	Write the fractions individually.
	$=$	$\displaystyle \frac{2\sqrt{108}}{\sqrt{27}}-\frac{7\sqrt{54}}{\sqrt{27}}$
	↓	Pull a $\sqrt{4}=2$ out of the first root.
	$=$	$\displaystyle \frac{4\sqrt{27}}{\sqrt{27}}-\frac{7\cdot\sqrt{2}\cdot\sqrt{27}}{\sqrt{27}}$
	↓	Shorten the fractions.
	$=$	$\displaystyle 4-7\sqrt{2}$

$\left(\sqrt2-\sqrt{18}\right)^2$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers

$\displaystyle \left(\sqrt{2}-\sqrt{18}\right)^2$	$=$
	↓	Apply the binomial formulas.
	$=$	$\displaystyle \sqrt{2}^2-2\cdot\sqrt{2}\cdot\sqrt{18}+\sqrt{18}^2$
	$=$	$\displaystyle 2-12+18$
	$=$	$\displaystyle 8$

$\displaystyle \left(\sqrt{2}-\sqrt{18}\right)^2$	$=$	$\displaystyle \left(\sqrt{2}-3\sqrt{2}\right)^2$
	↓	conclude
	$=$	$\displaystyle \left(-2\sqrt{2}\right)^2$
	$=$	$\displaystyle (-2)^2\cdot(\sqrt{2})^2$
	$=$	$\displaystyle 4\cdot2$
	$=$	$\displaystyle 8$

$\left(2\sqrt7-3\right)\left(1-\sqrt{28}\right)$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers

$\displaystyle \left(2\sqrt{7}-3\right)\left(1-\sqrt{28}\right)$	$=$
	↓	Pull the 2 out of the root as: $2\sqrt{7}=\sqrt{4\cdot7}$
	$=$	$\displaystyle \left(\sqrt{28}-3\right)\left(1-\sqrt{28}\right)$
	↓	Multiply out the brackets.
	$=$	$\displaystyle \sqrt{28}-28-3+3\sqrt{28}$
	↓	Simplify.
	$=$	$\displaystyle \sqrt{28}-31+3\sqrt{28}$
	$=$	$\displaystyle 4\sqrt{28}-31$

$3\sqrt{63}+6\sqrt{72}-4\sqrt{28}-17\sqrt8$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers

$\displaystyle 3\sqrt{63}+6\sqrt{72}-4\sqrt{28}-17\sqrt{8}$	$=$
	↓	Factorise the values under the root.
	$=$	$\displaystyle 3\sqrt{7\cdot9}+6\sqrt{2\cdot36}-4\sqrt{7\cdot4}-17\sqrt{2\cdot4}$
	↓	Take square roots.
	$=$	$\displaystyle 3\cdot3\sqrt{7}+6\cdot6\sqrt{2}-4\cdot2\sqrt{7}-34\sqrt{2}$
	$=$	$\displaystyle 9\sqrt{7}+36\sqrt{2}-8\sqrt{7}-34\sqrt{2}$
	↓	Conclude.
	$=$	$\displaystyle \sqrt{7}+2\sqrt{2}$

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