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