THE ROOT OF THE ROOTS

[vicslick46]

Hi, this was the explanation given for the problem CAT problem (simplifying the equation), I see how cube root 80 turns into cube root 16*5, but how does it go from that, to 12*root 5?


√(█(∛ 80 + 3/(9+4√ 5) ((9-4√ 5))/(9-4√ 5))
=

√(█(∛(16*5)+(27-12√5)/(81-(16*5))
=

√(12√5+27-12√5)

=

√27

[helloriteshranjan]

fishy step...!!
I mean wrong.

where does this problem appear?

[Ben Ku]

Hi, this was the explanation given for the problem CAT problem (simplifying the equation), I see how cube root 80 turns into cube root 16*5, but how does it go from that, to 12*root 5?

3√ 80 is not the cube root of 80. instead it's 3 * √ 80.

3 * √ 80
= 3 * √ (16*5)
= 3 * 4 * √ 5
= 12 * √ 5

Hope that helps.

[sadat.sathak]

The denominator of the second part under the root --> "81-(16)(5)" Seems to have dissapeared from the 2nd step onwards. What gives?

[jnelson0612]

The denominator of the second part under the root --> "81-(16)(5)" Seems to have dissapeared from the 2nd step onwards. What gives?

When you solve that part out, you get 81 - 80 = 1, so the denominator of 1 can be safely removed from the equation. :-)

Best wishes,

[hidden]

Following up on this question, what does it mean when the explanation says "To eliminate a radical in the denominator, we can multiply both the numerator and the denominator by the conjugate of that denominator"? Could someone perhaps refer me to the page in the MGMAT text which teaches this? I'm very confused about the rule, and how we eliminate that denominator. Thank you!

[jnelson0612]

Following up on this question, what does it mean when the explanation says "To eliminate a radical in the denominator, we can multiply both the numerator and the denominator by the conjugate of that denominator"? Could someone perhaps refer me to the page in the MGMAT text which teaches this? I'm very confused about the rule, and how we eliminate that denominator. Thank you!

Sure!
Take a look at Ben's post in this thread: fractions-simplifying-fractions-t7821.html He is showing how to multiply by the conjugate to get rid of the square root in the denominator of this problem. Remember, it's not okay to leave square roots in denominators. We have to get rid of them, so we multiply by the conjugate on the top and bottom of the fraction, since it's perfectly okay to have square roots in numerators.

Let us know if we can help you further!

[hidden]

Perfect thanks! My flashcard collection is getting so big...

:) At least I'm learning!

[tim]

glad to hear it!