
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
November 17th, 2017, 01:52 PM  #1 
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10  square both sides of an equation
Hello. With this question, would I solve what is in () first or would I see it as multiplication? I have not seen a problem with () in it with also a radical. Normally they just have me try to isolate the radical so it would be more like the problem without the () and I think my first step would be to move the 5 to the other side then divide out the 6, then take the square root of the both sides, etc. Thanks as always for any reply. Last edited by skipjack; November 17th, 2017 at 07:15 PM. 
November 17th, 2017, 03:13 PM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,966 Thanks: 807 
Your equation isn't complete! Is it supposed to be equal to 0? That is, is it supposed to be $\displaystyle (3x+ 4) 6\sqrt{3x+ 4}+ 5= 0$? The parentheses, $(3x+ 4)$, really have no meaning here. This is exactly the same as $\displaystyle 3x+ 4 6\sqrt{3x+ 4}+ 5= 0$ which can be reduced, since 4 + 5 = 9, to $\displaystyle 3x+ 9 6\sqrt{3x+ 4}= 0$. Now, what about that square root? We would like to get rid of it by squaring, but if we just squared as it is, we would get "mixed" terms that would still involve the square root. So the best thing to do is isolate that square root by adding $\displaystyle 6\sqrt{x+ 4}$ to both sides: $\displaystyle 3x+ 9= 6\sqrt{3x+ 4}$. Squaring both sides gives $\displaystyle (3x+ 9)^2= 9x^2+ 54x+ 81= 36(3x+ 4)= 108x+ 144$. That reduces to the quadratic equation $\displaystyle 9x^2 54x 60= 0$. Solve that quadratic equation for $x$. Check any solution you get, to that equation, in the original square root equation  squaring both sides of an equation can introduce "spurious solutions" that satisfy the new equation but not the first. After going back and correcting my typing errors, I noticed, finally, that we have $3x+ 4$ both inside and outside the square root. So we can let $\displaystyle y= \sqrt{3x+ 4}$ and have the equation $\displaystyle y^2 6y+ 5= (y 5)(y 1)= 0$. The roots to that are $y$ = 5 and $y$ = 1. So $\displaystyle y= \sqrt{3x+ 4}= 5$ gives $\displaystyle 3x+ 4= 25$, $\displaystyle 3x= 21$ and $\displaystyle x= 7$. $\displaystyle y= \sqrt{3x+ 4}= 1$ gives $\displaystyle 3x 4= 1$, $\displaystyle 3x= 5$, $\displaystyle x= \frac{5}{3}$. Again, you should check in the original equation to see whether they satisfy it. Last edited by skipjack; November 17th, 2017 at 07:41 PM. 
November 17th, 2017, 04:15 PM  #3 
Senior Member Joined: Aug 2012 Posts: 1,772 Thanks: 481  I'd start by moving the radical term to the right side so that it's by itself. Then when we square both sides it will be easier because there's only one term and the square root will go away.

November 17th, 2017, 07:40 PM  #4  
Global Moderator Joined: Dec 2006 Posts: 18,697 Thanks: 1525  Quote:
That can be solved by writing it as $9(x  7)(x + 1) = 0$, by using the quadratic formula or by completing the square. No, it gives $3x + 4= 1$, so $3x = 3$ (which means $x = 1$).  

Tags 
equation, sides, square 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Differentiate both sides of equation (labour demand and supply)  samuelhzb  Calculus  8  December 24th, 2016 11:09 PM 
Equation with x on both sides  SenatorArmstrong  Elementary Math  2  October 18th, 2016 01:34 PM 
Find length of sides of quadrilateral; write equation  David Dudek  Algebra  2  October 14th, 2012 06:49 PM 
Multiplying/dividing both sides of the equation..  badrush  Algebra  3  January 31st, 2010 03:51 PM 
Det. length of sides of polygon from Area and ratio of sides  telltree  Algebra  0  January 21st, 2010 12:51 PM 