I have a small Inter. Alg quiz coming up and I am stuck with 2 problems

[JoshRawrrs1]

I have a small Inter. Alg quiz coming up and I am stuck with 2 problems. One problem would be this

3) Rationalize the denominator. (All variables represent positive real numbers.)

a) \(\displaystyle \dfrac{7}{\sqrt[6]{8a^5 b^{14} c\,}}\)

I know you have to multiply the denom to both sides, but what I don't get is how to apply the 6th root on to both sides.

Also, I have a question in regarding to completing the square

The question is

"Solve the quadratic equation by completing the square"

5k^2 + 15k + 12 = 0

I subtract 12 to the other side making it

5k^2 + 15k = -12

I cannot divide by 3 or 5, I am stuck, my friend told me to divide 15 by 2, but how? You can't divide a single digit without dividing others to...

 

[HallsofIvy]

I have a small Inter. Alg quiz coming up and I am stuck with 2 problems. One problem would be this

3) Rationalize the denominator. (All variables represent positive real numbers.)

a) \(\displaystyle \dfrac{7}{\sqrt[6]{8a^5 b^{14} c\,}}\)

I know you have to multiply the denom to both sides, but what I don't get is how to apply the 6th root on to both sides.

There are no "two sides". Did you mean the numerator and denominator? But you don't "apply the 6th root on both sides" or both numerator and denominator. You take only the 6th root of the denominator as indicated.
To "rationalize the denominator" when you have a 6th root you multiply numerator and denominator by the 5/6 root:
\(\displaystyle \sqrt[6]{A}(\sqrt[6]{A})^5= (\sqrt[6]{A})^6= A\) or, equivalently \(\displaystyle \sqrt[6]{A}\sqrt[6]{A^5}= \sqrt[6]{A^6}= A\).

Also, I have a question in regarding to completing the square

The question is

"Solve the quadratic equation by completing the square"

5k^2 + 15k + 12 = 0

I subtract 12 to the other side making it

5k^2 + 15k = -12

I cannot divide by 3 or 5, I am stuck, my friend told me to divide 15 by 2, but how? You can't divide a single digit without dividing others to...

You certainly can divide both sides by 5: \(\displaystyle k^2+ k= -\frac{12}{5}\).
Now complete the square on the left.

 

[JoshRawrrs1]

There are no "two sides". Did you mean the numerator and denominator? But you don't "apply the 6th root on both sides" or both numerator and denominator. You take only the 6th root of the denominator as indicated.
To "rationalize the denominator" when you have a 6th root you multiply numerator and denominator by the 5/6 root:
\(\displaystyle \sqrt[6]{A}(\sqrt[6]{A})^5= (\sqrt[6]{A})^6= A\) or, equivalently \(\displaystyle \sqrt[6]{A}\sqrt[6]{A^5}= \sqrt[6]{A^6}= A\).

Thanks for the help on the second one. Yes On the first question I do mean the denom and numerator.

Edit: Does that mean that i multiply 7 to the denom? Making it:

\(\displaystyle 7\, \cdot\, \left(\sqrt[6]{8 a^5 b^{14} c\, }\right)^5\)

How would you go about solving this? Do I apply the 5 to everything?

 

[HallsofIvy]

Thanks for the help on the second one. Yes On the first question I do mean the denom and numerator.

Edit: Does that mean that i multiply 7 to the denom? Making it:

\(\displaystyle 7\, \cdot\, \left(\sqrt[6]{8 a^5 b^{14} c\, }\right)^5\)

How would you go about solving this? Do I apply the 5 to everything?

You, as I am sure you have learned before, multiply both numerator and denominator by what I gave before:

\(\displaystyle \left(\dfrac{7}{\sqrt[6]{8a^5b^{14}c\,}}\right)\,\left(\dfrac{\left(\sqrt[6]{8a^5b^{14}c\,}\right)^5}{{\left(\sqrt[6]{8a^5b^{14}c\,}\right)^5}}\right)\)\(\displaystyle \,=\, \dfrac{7\left(\sqrt[6]{8a^5b^{14}c\,}\right)^5}{8a^5b^{14}c}\)

 

[Deleted member 4993]

So you want to simplify 7 / (8 a^5 b^14 c)^(1/6)

No... the instruction was rationalize (a complicated proposal for hockey-players)

As far as simplification goes ... it is in the eye of the beholder. I think the problem stated the expression in its simplest form!!

 

[pka]

One problem would be this
\(\displaystyle \boxed{ \ \dfrac{7\sqrt[6]{8ab^4c^5}}{2ab^3c\; \ } \ }\) CORRECT

But why in the world would anyone go through so much "stuff" to get there.
It was noted the we want multiples of six as exponents change: \(\displaystyle 2^3 a^5 b^{14} c\) by multiplying \(\displaystyle 2^3 a b^{4} c^5\)
So \(\displaystyle \boxed{ \ \dfrac{7\sqrt[6]{8ab^4c^5}}{2ab^3c \ } \ }\).

Two steps at most (maybe three).

 

[lookagain]

3) Rationalize the denominator. (All variables represent positive real numbers.)

a) \(\displaystyle \dfrac{7}{\sqrt[6]{8a^5 b^{14} c\,}}\)

I'm going to answer it with simplified radicals. I will be multiplying the top and
bottom of the fraction by a radical, so that the radicand in the denominator
becomes the next higher-up sixth power.

\(\displaystyle \dfrac{7}{\sqrt[6]{8a^5b^{14}c \ } \ } \ = \)

\(\displaystyle \dfrac{7}{\sqrt[6]{(b^{12})8a^5b^2c \ } \ } \ = \)

\(\displaystyle \dfrac{7}{b^2\sqrt[6]{8a^5b^2c \ } \ } \ = \)

\(\displaystyle \dfrac{7}{b^2\sqrt[6]{2^3a^5b^2c^1 \ } \ } \cdot \ \dfrac{\sqrt[6]{2^3a^1b^4c^5 \ } \ }{\sqrt[6]{2^3a^1b^4c^5 \ } \ } \ =\)

\(\displaystyle \dfrac{7\sqrt[6]{8ab^4c^5}}{b^2\sqrt[6]{2^6a^6b^6c^6 \ } \ } \ = \)

\(\displaystyle \dfrac{7\sqrt[6]{8ab^4c^5}}{b^2(2abc) \ } \ = \)

\(\displaystyle \boxed{ \ \dfrac{7\sqrt[6]{8ab^4c^5}}{2ab^3c \ } \ }\)