anonymous
  • anonymous
Why is the cube root of a decimal number larger than the square root of the same decimal number? Let's say the number is 0.4225. It's square root is 0.65 while its cube root is 0.75037. Why is that?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
arindameducationusc
  • arindameducationusc
Are you there @emcrazy14 ?
ganeshie8
  • ganeshie8
maybe first think why the "cube of a number" is less than the "square of a number" when the number is between 0 and 1
ganeshie8
  • ganeshie8
or start here : if the number is between 0 and 1, why the "square of a number" is less than the "actual number" ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ganeshie8
  • ganeshie8
consider a concrete example : why \(\large 0.8^2\) is less than \(\large 0.8\) ?
anonymous
  • anonymous
@ganeshie8 : Okay, so the logic here is that since squaring or taking cube of a number between 0 & 1 yields a smaller number than the original number, in the same way the square root & the cube root tend to increase?
ganeshie8
  • ganeshie8
before that, do we see why squaring a number between 0 & 1 gives a smaller result ?
anonymous
  • anonymous
Yes. \[0.8^{2}\] is like solving for 80% of 0.8 which will definitely be less than 0.8.
ganeshie8
  • ganeshie8
Excellent! so from that how are we going to conclude that \(\large 0.8^3 \lt 0.8^2\) ?
anonymous
  • anonymous
Multiplying a number with a number smaller than 1 but great than 0 decreases it \[5 \times \frac{1}{2}=2.5<5\] The same is true for numbers smaller than 1 but greater than itself \[\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}<\frac{1}{2}\] Now suppose \[x=(\frac{1}{2})^6=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \] \[\sqrt{x}=\sqrt{(\frac{1}{2})^{6}}=(\frac{1}{2})^{3}=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\] and \[x^\frac{1}{3}=((\frac{1}{2})^6)^\frac{1}{3}=(\frac{1}{2})^2=\frac{1}{2} \times \frac{1}{2}\] Notice in square root, we are multiplying 3 times, so it will become ever smaller in cube root it is multiplied 2 times, so it will be as small
anonymous
  • anonymous
but greater than 0*
Michele_Laino
  • Michele_Laino
hint: I think that we have to specify if that number is greater than1 or less than 1. If: \[\Large \begin{gathered} x = \sqrt N \hfill \\ y = \sqrt[3]{N} \hfill \\ \end{gathered} \] then: \[\Large \begin{gathered} \log x = \frac{1}{2}\log N \hfill \\ \log y = \frac{1}{3}\log N \hfill \\ \end{gathered} \] so, if N>1, then \[\Large \log x > \log y \Rightarrow x > y\]
anonymous
  • anonymous
Wow. Thank you so much guys. It makes sense now. :) @Michele_Laino @Nishant_Garg @ganeshie8
Michele_Laino
  • Michele_Laino
:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.