MHB Solving for Real Solutions: x^2 − 10x + 1 = (x + 1)√x

[anemone]

Solve for real solutions of $x^2 − 10x + 1 = (x + 1)\sqrt{x}$.

 

[lfdahl]

My solution:

Spoiler

Squaring both sides and rearranging terms:$(x^2-10x+1)^2=((x+1)\sqrt{x})^2 \Rightarrow x^4-21x^3+100x-21x+1 = 0$
The last expression implies a product of the form: $(x^2+ax+1)(x^2+bx + 1)=0$
Multiplying and comparing coefficients:
$(x^2+ax+1)(x^2+bx + 1) = x^4 + (a+b)x^3+(2+a\cdot b)x^2+(a+b)x + 1=0$
It is easy to see, that $a = -7$ and $b = -14$ is the only pair of integers, which ensure the identity.
So, we have: $x^4-21x^3+100x-21x+1 = (x^2-7x+1)(x^2-14x+1) = 0$
The four positive roots to be evaluated are:\[\left \{ \frac{1}{2}(7\pm 3\sqrt{5}), 7\pm 4\sqrt{3} \right \}\]Both roots: $\left \{ \frac{1}{2}(7 + 3\sqrt{5}), \frac{1}{2}(7 - 3\sqrt{5})\right \}$ imply, that $x^2-10x+1 < 0$, but the RHS: $(x+1)\sqrt{x} > 0$. These roots would be valid, if we had the squared version ($4th$ degree polynomial) as starting point.The two other roots preserve the plus sign in $x^2-10x+1$, and therefore, $\left \{ 7\pm 4\sqrt{3}\right \}$ are the only valid roots.

 

[anemone]

Well done lfdahl, and thanks for participating!(Cool)

My solution:

Spoiler

Rewrite the original equation as

$(x+1)^2-12x = \sqrt{(x + 1)^2}\sqrt{x}$

Next, let $u=(x+1)^2$ then the equation above turns into $u-12x=\sqrt{u}\sqrt{x}$, or $ u-\sqrt{u}\sqrt{x}-12x=0$, which can be factorized into $(\sqrt{u}-4\sqrt{x}) (\sqrt{u}+3\sqrt{x})=0$, which suggests $u=16x$ and back substitute $u=(x+1)^2$ into $u=16x$ yields $x^2-14x+1=0$, i.e. $x=7\pm 4\sqrt{3}$.