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Indices and Roots

Questions based on Exercise Intro(b) from the Textbook \"Engineering Maths through Applications\" by Kuldeep Singh.

Intended for use as Homework Exercises.

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Evaluate:

In these questions order does not matter. It is often easier to take roots inside and then do the multiplication or division

\n

a)   $\\sqrt{\\var{a1} \\times \\var{a2}}=\\sqrt{\\var{a1}} \\times \\sqrt{\\var{a2}}=\\var{a1a} \\times \\var{a2a}=\\var{a3}$

\n

b)   $\\sqrt{\\var{b1} \\times \\var{b2}}=\\sqrt{\\var{b1}} \\times \\sqrt{\\var{b2}}=\\var{b1a} \\times \\var{b2a}=\\var{b3}$

\n

The same applies to division:

\n

c)   $\\sqrt{\\frac{144}{\\var{c3}}}=\\frac{\\sqrt{144}}{\\sqrt{\\var{c3}}}=\\frac{\\var{c1a}}{\\var{c3a}}=\\var{c4}$

\n

d)   $\\sqrt{\\frac{225}{25}}=\\frac{\\sqrt{225}}{\\sqrt{25}}=\\frac{15}{5}=3$

\n

For e) and f) the numbers/roots given are not so easy to identify - after all, who memorises 5th roots? So better to use your calculator (with care). But the above method still works:

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e)   $\\sqrt[5]{\\frac{7776}{243}}=\\frac{\\sqrt[5]{7776}}{\\sqrt[5]{243}}=\\frac{6}{3}=2$

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f)   Remember to use brackets to get the correct answer from your calculator, $=-2$

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$\\sqrt{\\frac{144}{\\var{c3}}}=$[[0]]

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$\\sqrt{\\frac{225}{\\var{25}}}=$[[0]]

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$\\sqrt[5]{\\frac{7776}{243}}=$[[0]]

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$\\sqrt[5]{-(\\frac{7776}{243})}=$[[0]]

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