
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
October 1st, 2017, 11:13 AM  #1 
Member Joined: Sep 2014 From: Sweden Posts: 75 Thanks: 0  Simpler solve for this equation: sqrt(x^2x10)sqrt(x^211x)=10
Hi, I have solved this equation but it seems like there might be a simpler way to do it (or it is not a simpler way at all ), but however the equation that I have solved looks like this: $\displaystyle \displaystyle \sqrt{x^2x10}\sqrt{x^211x}=10\\\\\\\\ \displaystyle \sqrt{x^2x10}=10+\sqrt{x^211x}\\\\\\\\ \displaystyle x^2x10=100+20\sqrt{x^211x}+x^211x\\\\\\\\ \displaystyle 10x20\sqrt{x^211x}=110\\\\\\\\ \displaystyle 10(x2\sqrt{x^211x})=11\\\\\\\\ \displaystyle x11=2\sqrt{x^211x}\\\\\\\\ \displaystyle x^222x+121=4(x^211)\\\\\\\\ \displaystyle x^222x+121=4x^244x\\\\\\\\ \displaystyle 3x^2+22x+121=0\\\\\\\\ $ Use the abcformula $\displaystyle \displaystyle x=\frac{b\pm\sqrt{b^24ac}}{2a}\\\\\\\\$ $\displaystyle \displaystyle x=\frac{22\pm\sqrt{(22)^24\cdot 3\cdot 121}}{2\cdot 3}\\\\\\\\ \displaystyle x=\frac{22\pm\sqrt{484+1452}}{6}\\\\\\\\ \displaystyle x=\frac{22\pm\sqrt{1936}}{6}\\\\\\\\ \displaystyle x=\frac{22\pm44}{6}\\\\\\\\ $ The first solution (false) $\displaystyle \displaystyle x_1=\frac{22+44}{6}\\\\\\\\ \displaystyle x_1=\frac{22}{6}\\\\\\\\ \displaystyle x_1=\frac{11}{3} \ false\\\\\\\\ $ The second solution (true) $\displaystyle \displaystyle x_2=\frac{2244}{6}\\\\\\\\ \displaystyle x_2=\frac{66}{6}\\\\\\\\ \displaystyle x_2=\frac{66}{6}\\\\\\\\ \displaystyle x_2=11 \ true\\\\\\\\ $ Last edited by DecoratorFawn82; October 1st, 2017 at 11:16 AM. 
October 1st, 2017, 01:35 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,454 Thanks: 567 
There is no simpler way. Squaring steps introduce spurious roots. Fourth line has an error, but it was corrected for the fifth line.

October 1st, 2017, 06:03 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,704 Thanks: 1529 
Multiplying both sides by $\sqrt{x^2  x  10} + \sqrt{x^2  11x} + 10$, gathering like terms and dividing by 10 leads to $x  11 = 2\sqrt{x^2  11x}$, so $x^2  22x + 121 = 4x^2  44x$. Hence $3x^2  22x + 121 = 0$, i.e. $(x  11)(3x + 11) = 0$. As $x = 11/3$ doesn't satisfy the original equation, $x = 11$ is the unique solution. 

Tags 
equation, simpler, solve, sqrtx2x10sqrtx211x10 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Consider the demand equation p = sqrt((200x)/(2x))  Lalaluye  Calculus  2  July 3rd, 2017 07:52 PM 
How can I solve this Integration sqrt(x^2+4)  Married Math  Calculus  7  September 19th, 2014 07:56 AM 
Easier way to solve lim_{x > infty} (sqrt(9x^2x)3x)  Singularity  Calculus  4  October 24th, 2012 11:44 AM 
X=sqrt(Xsqrd+Ysqrd)+AB; Solve for X?  MathJustForFun  Algebra  3  May 13th, 2010 01:03 AM 
simplifying sqrt[a + sqrt(b)]  nikkor180  Real Analysis  3  June 25th, 2009 03:18 AM 