Eirik's copy of Solve equations which include a single root (e.g. \sqrt{x}=blah)

$\begingroup$

Question 1

Please complete the following.

a)

If $\sqrt[\var{intpower}]{x}=\var{intrhs}$ , then $x=$ interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{4}$ .

Score: 0/1

Show feedback

This feedback is based on your last submitted answer. Save your changed answer to get updated feedback.

Or, you could:

There's nothing more to do from here.

b)

If $\simplify{{bxcoeff}y^(1/{bpower})+{bb}}=\var{bc}$ , then $y=$ interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{81}$ .

Score: 0/1

Show feedback

This feedback is based on your last submitted answer. Save your changed answer to get updated feedback.

Or, you could:

There's nothing more to do from here.

c)

For this question, if the answer was $\left(\frac{35}{11}\right)^{11}-24$ , then you could enter (35/11)^(11)-24.

If $\displaystyle{\simplify{(root(z+{db},{dpower}))/{ddenom}}}=\var{dc}$ , then $z=$ interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{468^{ 6 } + 65}$ .

Score: 0/1

Show feedback

This feedback is based on your last submitted answer. Save your changed answer to get updated feedback.

Or, you could:

There's nothing more to do from here.

Advice

a) Given $\sqrt[\var{intpower}]{x}=\var{intrhs}$ , we raise both sides to the power of $\var{intpower}$ to get $x$ by itself.

$\sqrt[\var{intpower}]{x}$	$=$	$\var{intrhs}$

$\left(\sqrt[\var{intpower}]{x}\right)^{\var{intpower}}$	$=$	$\simplify[basic]{({intrhs})^{intpower}}$

$x$	$=$	$\var{intsoln}$

b) Given $\simplify{{bxcoeff}y^(1/{bpower})+{bb}}=\var{bc}$ , we can rearrange the equation to get $y^\frac{1}{\var{bpower}}$ by itself and then we can raise both sides to the power of $\var{bpower}$ to get $y$ by itself.

$\simplify{{bxcoeff}y^(1/{bpower})+{bb}}$	$=$	$\var{bc}$

$\simplify{{bxcoeff}y^(1/{bpower})}$	$=$	$\simplify[basic]{{bc}-{bb}}$

$\simplify{{bxcoeff}y^(1/{bpower})}$	$=$	$\simplify{{bc-bb}}$

$y^\frac{1}{\var{bpower}}$	$=$	$\simplify[!basic]{{bc-bb}/{bxcoeff}}$

$y^\frac{1}{\var{bpower}}$	$=$	$\simplify{{bc-bb}/{bxcoeff}}$

$\left(y^\frac{1}{\var{bpower}}\right)^{\var{bpower}}$	$=$	$\simplify[basic]{({(bc-bb)/bxcoeff})^{bpower}}$

$y$	$=$	$\var{bsoln}$

c) Given $\displaystyle{\simplify{(root(z+{db},{dpower}))/{ddenom}}}=\var{dc}$ , we can rearrange the equation to get $\simplify{(root(z+{db},{dpower}))}$ by itself, then we can raise both sides to the power of $\var{dpower}$ , and finally rearrange to get $z$ by itself.

$\displaystyle{\simplify{(root(z+{db},{dpower}))/{ddenom}}}$	$=$	$\var{dc}$

$\displaystyle{\simplify{(root(z+{db},{dpower}))}}$	$=$	$\simplify[basic]{{dc}*{ddenom}}$

$\displaystyle{\simplify{(root(z+{db},{dpower}))}}$	$=$	$\var{dc*ddenom}$

$\left(\sqrt[\var{dpower}]{\simplify{z+{db}}}\right)^\var{dpower}$	$=$	$\simplify[basic]{({dc*ddenom})^{dpower}}$

$\simplify{z+{db}}$	$=$	$\simplify[basic]{({(dc*ddenom)})^{dpower}}$

$z$	$=$	$\simplify[basic]{({(dc*ddenom)})^{dpower}-{db}}$

$\endgroup$

Numbas has failed

Eirik's copy of Solve equations which include a single root (e.g. \sqrt{x}=blah)

a)

b)

c)

Advice