Algebraic Verification of Radical Equations: Solving Without a Calculator

[mathdad]

Verify that both sides of the radical equation agree without using a calculator. See picture. How can this be done algebraically?

View attachment 7968

 

[MarkFL]

I would first observe that $1+\sqrt{5}$ is a root of:

\(\displaystyle x^2-2x-4=0\)

And so, the coefficients of the expansion:

\(\displaystyle (1+\sqrt{5})^n\)

Can be found recursively via:

\(\displaystyle A_{n}=2A_{n-1}+4A_{n-2}\)

For the rational term, we have:

\(\displaystyle A_0=1,\,A_1=1\)

Hence:

\(\displaystyle A_2=2(1)+4(1)=6\)
\(\displaystyle A_3=2(6)+4(1)=16\)
\(\displaystyle A_4=2(16)+4(6)=56\)
\(\displaystyle A_5=2(56)+4(16)=176\)

And for the irrational term, we have:

\(\displaystyle A_0=0,\,A_1=1\)

\(\displaystyle A_2=2(1)+4(0)=2\)
\(\displaystyle A_3=2(2)+4(1)=8\)
\(\displaystyle A_4=2(8)+4(2)=24\)
\(\displaystyle A_5=2(24)+4(8)=80\)

And so we may conclude:

\(\displaystyle (1+\sqrt{5})^5=176+80\sqrt{5}\)

And the result follows. :)

 

[mathdad]

I would first observe that $1+\sqrt{5}$ is a root of:

\(\displaystyle x^2-2x-4=0\)

And so, the coefficients of the expansion:

\(\displaystyle (1+\sqrt{5})^n\)

Can be found recursively via:

\(\displaystyle A_{n}=2A_{n-1}+4A_{n-2}\)

For the rational term, we have:

\(\displaystyle A_0=1,\,A_1=1\)

Hence:

\(\displaystyle A_2=2(1)+4(1)=6\)
\(\displaystyle A_3=2(6)+4(1)=16\)
\(\displaystyle A_4=2(16)+4(6)=56\)
\(\displaystyle A_5=2(56)+4(16)=176\)

And for the irrational term, we have:

\(\displaystyle A_0=0,\,A_1=1\)

\(\displaystyle A_2=2(1)+4(0)=2\)
\(\displaystyle A_3=2(2)+4(1)=8\)
\(\displaystyle A_4=2(8)+4(2)=24\)
\(\displaystyle A_5=2(24)+4(8)=80\)

And so we may conclude:

\(\displaystyle (1+\sqrt{5})^5=176+80\sqrt{5}\)

And the result follows. :)

What if I decided to raise both sides to the 5th power? Can it be done this way as well?

 

[Wilmer]

What if I decided to raise both sides to the 5th power? Can it be done this way as well?

Well, you'd have quite a few terms to manipulate; as example:

(a + b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5

 

[mathdad]

Well, you'd have quite a few terms to manipulate; as example:

(a + b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5

I am not familiar with the method you introduced here involving the root of the quadratic equation.

 

[Wilmer]

Didn't "introduce" anything...
YOU asked about raising to 5th power...
Gave you an example.
HOKAY?!

 

[mathdad]

Didn't "introduce" anything...
YOU asked about raising to 5th power...
Gave you an example.
HOKAY?!

1. This conversation is between Mark and me.

2. I would like for you to stop commenting in my posts. To you everything is a joke.

 

[Wilmer]

2. I would like for you to stop commenting in my posts. To you everything is a joke.

Will do; pleasure is all mine. All yours Mark...

 

[mathdad]

Thank you very much, Mark. Interesting notes as always...

 

[MarkFL]

What if I decided to raise both sides to the 5th power? Can it be done this way as well?

Yes, and as Wilmer was pointing out, you would likely want to use the binomial theorem. If you raise both sides to the 5th power, you get:

\(\displaystyle 176+80\sqrt{5}=(1+\sqrt{5})^5\)

Now, using the binomial theorem on the RHS, we obtain:

\(\displaystyle 176+80\sqrt{5}=1+5\cdot5^{\Large\frac{1}{2}}+10\cdot5^{\Large\frac{2}{2}}+10\cdot5^{\Large\frac{3}{2}}+5\cdot5^{\Large\frac{4}{2}}+5^{\Large\frac{5}{2}}\)

\(\displaystyle 176+80\sqrt{5}=1+5\sqrt{5}+10\cdot5+10\cdot5\sqrt{5}+5\cdot5^2+5^2\sqrt{5}\)

\(\displaystyle 176+80\sqrt{5}=1+5\sqrt{5}+50+50\sqrt{5}+125+25\sqrt{5}\)

\(\displaystyle 176+80\sqrt{5}=176+80\sqrt{5}\)

This is an identity, and so we know the original equation is true.

 

[mathdad]

Mark:

Another amazing reply!