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 Algebra Pre-Algebra and Basic Algebra Math Forum

November 17th, 2017, 01:52 PM   #1
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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.
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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: 3,264 Thanks: 902 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. Thanks from GIjoefan1976 Last edited by skipjack; November 17th, 2017 at 07:41 PM. November 17th, 2017, 04:15 PM   #3
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Quote:
 Originally Posted by GIjoefan1976 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
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
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Quote:
 Originally Posted by Country Boy That reduces to the quadratic equation $\displaystyle 9x^2- 54x- 60= 0$.
No, it reduces to $\displaystyle 9x^2- 54x- 63= 0$.

That can be solved by writing it as $9(x - 7)(x + 1) = 0$, by using the quadratic formula or by completing the square.

Quote:
 Originally Posted by Country Boy $\displaystyle y= \sqrt{3x+ 4}= 1$ gives $\displaystyle 3x- 4= 1$, $\displaystyle 3x= 5$, $\displaystyle x= \frac{5}{3}$.
No, it gives $3x + 4= 1$, so $3x = -3$ (which means $x = -1$). Tags equation, sides, square Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post samuelhzb Calculus 8 December 24th, 2016 11:09 PM SenatorArmstrong Elementary Math 2 October 18th, 2016 01:34 PM David Dudek Algebra 2 October 14th, 2012 06:49 PM badrush Algebra 3 January 31st, 2010 03:51 PM telltree Algebra 0 January 21st, 2010 12:51 PM

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