Welcome to our community

Be a part of something great, join today!

Problem Of The Week #425 July 13th, 202

Status
Not open for further replies.
  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,632
Here is this week's POTW:

-----

Solve the equation $\dfrac{7}{\sqrt{x^2-10x+26}+\sqrt{x^2-10x+29}+\sqrt{x^2-10x+41}}=x^4-9x^3+16x^2+15x+26$

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Admin
  • #2

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,632
Congratulations to Opalg (Cool) for his correct solution, which you can find below:

Solution from Opalg :
Since $x^2-10x + 26 = (x-5)^2 + 1$, $x^2-10x + 29 = (x-5)^2 + 4$ and $x^2-10x + 41 = (x-5)^2 + 16$, it seems natural to wite the right side of the equation in terms of $x-5$: $$\begin{aligned} x^4-9x^3+16x^2+15x+26 &= (x-5)(x^3 - 4x^2 - 4x -5) + 1 \\ &= (x-5)^2(x^2 + x + 1) + 1. \end{aligned}$$ Therefore $\dfrac{7}{\sqrt{x^2-10x+26}+\sqrt{x^2-10x+29}+\sqrt{x^2-10x+41}}=x^4-9x^3+16x^2+15x+26$ can be written as $$\dfrac{7}{\sqrt{ (x-5)^2 + 1}+\sqrt{ (x-5)^2 + 4}+\sqrt{ (x-5)^2 + 16 }} = (x-5)^2(x^2 + x + 1) + 1.$$ If $x=5$ then both sides are equal to $1$. If $x\ne5$ then the left side is less than $1$ (because the denominator will be greater than $7$), and the right side is greater than $1$ (because $x^2+x+1$ is always positive). So $x=5$ is the only solution.
 
Status
Not open for further replies.