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

Intended for \nuse as Homework Exercises.

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$\\var{abc1}^{\\var{P1}}=$[[0]]

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$\\var{abc1}^{\\var{P2}}=$[[0]]

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$\\var{abc1}^{\\var{P3}}=$[[0]]

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$\\var{d1}^{\\var{d2}}=$[[0]]

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$\\var{e1}^{2}=$[[0]]

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$\\var{f1}^{0}=$[[0]]

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$(\\var{g1})^{\\var{g2}}=$[[0]]

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a)   $\\var{abc1}^{\\var{P1}}= \\var{abc1} \\times\\var{abc1} \\times... (\\var{P1} \\text{ times}) = \\var{a}$

\n

b)   $\\var{abc1}^{\\var{P2}}= \\var{abc1} \\times\\var{abc1} \\times... (\\var{P2} \\text{ times}) =\\var{b}$

\n

c)   $\\var{abc1}^{\\var{P3}} = \\var{abc1} \\times\\var{abc1} \\times... (\\var{P3} \\text{ times}) =\\var{c}$

\n

d)   $\\var{d1}^{\\var{d2}} = \\var{d1} \\times\\var{d1} \\times... (\\var{d2} \\text{ times}) =\\var{d}$

\n

e)  $\\var{e1}^{2}=\\var{e1} \\times \\var{e1} =\\var{e2}$

\n

f)   Any number to the power of  $0$  is equal to  $1$,  so  $\\var{f1}^{0}=1$

\n

g)   $(\\var{g1})^{\\var{g2}}=\\var{g}$

\n

Notice that an odd power of a negative number gives a negative answer.

", "statement": "

Evaluate the following:

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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.Indices and Roots

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a)   Since  $\\var{a1} \\times \\var{a1}=\\var{a2}$  and also $(-\\var{a1}) \\times (-\\var{a1})=\\var{a2}$,  so:

\n

\$\\pm \\sqrt{\\var{a2}}=\\pm \\var{a1}\$

\n

\n

\$\\pm \\sqrt{\\var{b2}}=\\pm \\var{b1}\$

\n

c)   Similarly:

\n

\$\\sqrt{\\var{c2}}=\\var{c1}\$

\n

d)   Your calculator will have a  $\\sqrt[3]{}$  button. Use it to evaluate:

\n

\$\\sqrt[3]{\\var{d2}}=\\var{d1}\$

\n

e)

\n

\$\\sqrt[3]{\\var{e2}}=\\var{e1}\$

\n

f)   Since we need the 4th root (even) of $\\var{f2}$, we will have two numbers. Using your calculator:

\n

\$\\pm \\sqrt[4]{\\var{f2}}=\\pm \\var{f1}\$

\n

g)

\n

\$\\pm \\sqrt[4]{\\var{g2}}=\\pm \\var{g1}\$

\n

h)   Again, the calculator shows the 8th root (even) of $\\var{h2}$ as  $\\var{h1}$. Are there any other roots? Yes, $-\\var{h1}$. Hence:

\n

\$\\pm \\sqrt[8]{\\var{h2}}=\\pm \\var{h1}\$

\n

i)   Using the calculator gives   $\\sqrt[5]{\\var{i2}}=\\var{i1}$

\n

j)   Since we are interested in the 5th root (odd), there can only be one answer:

\n

\$\\sqrt[5]{(\\var{j2})}=\\var{j1}\$

", "statement": "

Using your calculator or otherwise, compute:

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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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$\\sqrt[3]{\\var{d2}}=$[[0]]

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$\\pm \\sqrt[4]{\\var{f2}}=\\pm$[[0]]

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$\\pm \\sqrt[8]{\\var{h2}}=\\pm$[[0]]

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$\\sqrt[5]{\\var{i2}}=$[[0]]

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$\\sqrt[5]{(\\var{j2})}=$[[0]]

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$\\sqrt{\\var{a4}-\\var{a3}}=$[[0]]

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$\\sqrt{\\var{b4}+\\var{b3}}=$[[0]]

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$\\sqrt{\\var{c4}-\\var{c3}}=$[[0]]

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$\\sqrt[4]{\\var{d4}-\\var{d3}}=$[[0]]

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$\\sqrt[3]{\\var{e3}+\\var{e3}}=$[[0]]

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$\\sqrt[3]{(-\\var{e3}-\\var{e3})}=$[[0]]

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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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Using your calculator or otherwise, calculate:

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b1

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For these questions, we first need to carry out the arithmetic then take the root of the result:

\n

a)   $\\sqrt{\\var{a4}-\\var{a3}}=\\sqrt{\\var{a2}}=\\var{a1}$

\n

b)   $\\sqrt{\\var{b4}+\\var{b3}}=\\sqrt{\\var{b2}}=\\var{b1}$

\n

c)   $\\sqrt{\\var{c4}-\\var{c3}}=\\sqrt{\\var{c2}}=\\var{c1}$

\n

d)   $\\sqrt[4]{\\var{d4}-\\var{d3}}=\\sqrt[4]{\\var{d2}}=\\var{d1}$

\n

e)   $\\sqrt[3]{\\var{e3}+\\var{e3}}=\\sqrt[3]{\\var{e2}}=\\var{e1}$

\n

f)   $\\sqrt[3]{(-\\var{e3}-\\var{e3})}=\\sqrt[3]{-\\var{e2}}=-\\var{e1}$

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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:

\n

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

\n

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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c1a

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