1. ## Irrational equation solving

Hello,
How to solve this irrational equation$(x^2+4)^{1/3}=\sqrt{x-1}+2x-3$ ?

2. ## Re: Irrational equation solving

Originally Posted by Vinod
Hello,
How to solve this irrational equation$(x^2+4)^{1/3}=\sqrt{x-1}+2x-3$ ?
Inspection gives me x = 2. Otherwise unless you can pull a trick out of your hat you will have to cube both sides of the equation. Get some coffee...it's a long calculation.

-Dan

3. ## Re: Irrational equation solving

Let $u^2 = x-1$. This means $x = u^2+1$. Subbing in:

$$((u^2+1)^2+4)^{1/3} = u+2u^2-1$$

Cubing both sides gives:

$$(u^2+1)^2+4 = (2u^2+u-1)^3$$

$$u^4+2u^2+5 = 8u^6+12u^5-6u^4-11u^3+3u^2+3u-1$$

$$8u^6+12u^5-7u^4-11u^3+u^2+3u-6 = 0$$

$$(u-1)(8u^5+20u^4+13u^3+2u^2+3u+6) = 0$$

One solution is $u=1$. There are no other rational roots (using the Rational Root Theorem, you can find that there are no other rational roots).

So, $x = u^2+1=1^2+1=2$. Let's plug that in and see:

$$2 = (2^2+4)^{1/3} = \sqrt{2-1}+2(2)-3 = 2$$

This shows that $x=2$ is a solution.

4. ## Re: Irrational equation solving

Originally Posted by SlipEternal
Let $u^2 = x-1$. This means $x = u^2+1$. Subbing in:

$$((u^2+1)^2+4)^{1/3} = u+2u^2-1$$

Cubing both sides gives:

$$(u^2+1)^2+4 = (2u^2+u-1)^3$$

$$u^4+2u^2+5 = 8u^6+12u^5-6u^4-11u^3+3u^2+3u-1$$

$$8u^6+12u^5-7u^4-11u^3+u^2+3u-6 = 0$$

$$(u-1)(8u^5+20u^4+13u^3+2u^2+3u+6) = 0$$

One solution is $u=1$. There are no other rational roots (using the Rational Root Theorem, you can find that there are no other rational roots).

So, $x = u^2+1=1^2+1=2$. Let's plug that in and see:

$$2 = (2^2+4)^{1/3} = \sqrt{2-1}+2(2)-3 = 2$$

This shows that $x=2$ is a solution.
Hello,
On some math site, This equation has been solved as below.
$(x^2+4)^{1/3}=\sqrt{x-1}+2x-3$
$((x^2+4)^{1/3}-2)=(\sqrt{x-1}-1)+2x-6$ 1st step

$\Leftrightarrow\frac{x^2+4-8}{(x^2+4)^{2/3}+2(x^2+4)^{1/3}+4}=\frac{x-2}{\sqrt{x-1}+1}+2(x-2)$ 2nd step

$\Leftrightarrow(x-2)\frac{x+2}{(x^2+4)^{2/3}+2(x^2+4)^{1/3}+4}-\frac{1}{\sqrt{x-1}+1}-2=0$

(x-2)=0$\Leftrightarrow x=2$

Now in this answer 1st step of rationalising, and in the 2nd step left side of rationalisation are difficult to understand. Would you explain me those steps with calculations?

5. ## Re: Irrational equation solving

Originally Posted by Vinod
Hello,
On some math site, This equation has been solved as below.
$(x^2+4)^{1/3}=\sqrt{x-1}+2x-3$
$((x^2+4)^{1/3}-2)=(\sqrt{x-1}-1)+2x-6$ 1st step

$\Leftrightarrow\frac{x^2+4-8}{(x^2+4)^{2/3}+2(x^2+4)^{1/3}+4}=\frac{x-2}{\sqrt{x-1}+1}+2(x-2)$ 2nd step

$\Leftrightarrow(x-2)\frac{x+2}{(x^2+4)^{2/3}+2(x^2+4)^{1/3}+4}-\frac{1}{\sqrt{x-1}+1}-2=0$

(x-2)=0$\Leftrightarrow x=2$

Now in this answer 1st step of rationalising, and in the 2nd step left side of rationalisation are difficult to understand. Would you explain me those steps with calculations?
The idea was they knew that $a^3-b^3 = (a-b)(a^2+ab+b^2)$. So, they used that fact to say that $a-b = \dfrac{a^3-b^3}{a^2+ab+b^2}$. They just needed to set up a difference of cubes on the LHS. I am not sure why they chose 2. It appears they already knew that 2 was a solution, and they decided to use 2 for that reason. You have the 1st step wrong (it should be 2x-4 at the end, not -6).

Note that each denominator must be positive, which negates the possibility of division by zero errors (which is what allows them to factor out the $x-2$ term).

6. ## Re: Irrational equation solving

Originally Posted by Vinod
Hello,
On some math site, This equation has been solved as below.
$(x^2+4)^{1/3}=\sqrt{x-1}+2x-3$
$((x^2+4)^{1/3}-2)=(\sqrt{x-1}-1)+2x-6$ 1st step

$\Leftrightarrow\frac{x^2+4-8}{(x^2+4)^{2/3}+2(x^2+4)^{1/3}+4}=\frac{x-2}{\sqrt{x-1}+1}+2(x-2)$ 2nd step

$\Leftrightarrow(x-2)\frac{x+2}{(x^2+4)^{2/3}+2(x^2+4)^{1/3}+4}-\frac{1}{\sqrt{x-1}+1}-2=0$

(x-2)=0$\Leftrightarrow x=2$

Now in this answer 1st step of rationalising, and in the 2nd step left side of rationalisation are difficult to understand. Would you explain me those steps with calculations?
A small typo. The penultimate line should read:

$\displaystyle (x-2) \left [ \frac{x+2}{(x^2+4)^{2/3}+2(x^2+4)^{1/3}+4}-\frac{1}{\sqrt{x-1}+1}-2 \right ] =0$

-Dan