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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: 3,261 Thanks: 894 
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: 2,052 Thanks: 588  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: 19,731 Thanks: 1808  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 $x = 1$).  

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