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Chapter 1: Problem 3

Denest the following, if possible. a. \(\sqrt{3-2 \sqrt{2}}\) b. \(\sqrt{1+\sqrt{2}}\) c. \(\sqrt{5+2 \sqrt{6}}\) d. \(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\) e. Find the roots of \(x^{2}+6 x-4 \sqrt{5}=0\) in simplified form.

Short Answer

Expert verified
Part (a): \(\sqrt{3-2\sqrt{2}} = \sqrt{2}-1\), Part (b): \(\sqrt{1+\sqrt{2}} = 2\), Part (c): \(\sqrt{5+2\sqrt{6}} = 3\), Part (d): \(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}=\sqrt{5}\), Part (e): The roots of the equation are \(-3+\sqrt{9+4\sqrt{5}}\) and \(-3-\sqrt{9+4\sqrt{5}}\).

Step by step solution

01

Part (a)

The general approach used to denest square roots, in this format, is to assume \(\sqrt{3-2 \sqrt{2}}=a-b\). Squaring both sides gives \(3-2\sqrt{2}=a^{2}+b^{2}-2 a b\). This needs to look like the equation \(3-2 \sqrt{2}=a^{2}+b^{2}-\sqrt{2}a b * 2\). When we compare these two equations, it can be inferred that \(a=\sqrt{2}, b=1\). So, \(\sqrt{3-2 \sqrt{2}}= \sqrt{2}-1\)
02

Part (b)

As in Part (a), let's assume \(\sqrt{1+\sqrt{2}}=a+b\). Squaring both sides gives \(1+\sqrt{2}=a^{2}+b^{2}+2ab\). Comparing to \(1+\sqrt{2}=a^{2}+b^{2}+2a b\), it can be inferred that \(a=1, b=1\). So, \(\sqrt{1+\sqrt{2}}=1+1=2\)
03

Part (c)

Assume \(\sqrt{5+2 \sqrt{6}}=a+b\). By squaring the equation we get \(5+2\sqrt{6}=a^{2}+b^{2}+2ab\). It should be similar to the equation \(5+2\sqrt{6}=a^{2}+b^{2}+2a b\). By comparing these two equations we can see that \(a=2, b=1\). So, \(\sqrt{5+2 \sqrt{6}}=2+1=3\)
04

Part (d)

Given \(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\), change to \((x+2)^2=(x-2)^2=5\). From this equation we can find that \(x=\sqrt{5}\). Therefore \(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}=\sqrt{5}\)
05

Part (e)

For the quadratic equation given, we apply the quadratic formula with \(a=1\), \(b=6\), and \(c=-4\sqrt{5}\). The roots are given by \(-b\pm \sqrt{b^{2}-4ac}\div2a\). Thus, the roots are \(-6\pm\sqrt{36 - 4*(-4\sqrt{5})}/2=-6\pm \sqrt{36 + 16\sqrt{5}}/2=-3\pm\sqrt{9 + 4\sqrt{5}}\).

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