MHB Algebraic Verification of Radical Equations: Solving Without a Calculator

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

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

$$x^2-2x-4=0$$

And so, the coefficients of the expansion:

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

Can be found recursively via:

$$A_{n}=2A_{n-1}+4A_{n-2}$$

For the rational term, we have:

$$A_0=1,\,A_1=1$$

Hence:

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

And for the irrational term, we have:

$$A_0=0,\,A_1=1$$

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

And so we may conclude:

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

And the result follows. :)
 
MarkFL said:
I would first observe that $1+\sqrt{5}$ is a root of:

$$x^2-2x-4=0$$

And so, the coefficients of the expansion:

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

Can be found recursively via:

$$A_{n}=2A_{n-1}+4A_{n-2}$$

For the rational term, we have:

$$A_0=1,\,A_1=1$$

Hence:

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

And for the irrational term, we have:

$$A_0=0,\,A_1=1$$

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

And so we may conclude:

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

And the result follows. :)

What if I decided to raise both sides to the 5th power? Can it be done this way as well?
 
RTCNTC said:
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
 
Wilmer said:
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.
 
Last edited:
Didn't "introduce" anything...
YOU asked about raising to 5th power...
Gave you an example.
HOKAY?!
 
Wilmer said:
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.
 
RTCNTC said:
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...
 
Thank you very much, Mark. Interesting notes as always...
 
  • #10
RTCNTC said:
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:

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

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

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

$$176+80\sqrt{5}=1+5\sqrt{5}+10\cdot5+10\cdot5\sqrt{5}+5\cdot5^2+5^2\sqrt{5}$$

$$176+80\sqrt{5}=1+5\sqrt{5}+50+50\sqrt{5}+125+25\sqrt{5}$$

$$176+80\sqrt{5}=176+80\sqrt{5}$$

This is an identity, and so we know the original equation is true.
 
  • #11
Mark:

Another amazing reply!
 

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