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November 1, 2014

November 1, 2014

Posted by **Ashley** on Sunday, September 6, 2009 at 8:14pm.

Normally, I either take the square root of both sides or I square both sides. Since one side of the equation is square root of 3x-4 and the other side is (x-4) squared - 3, I am confused on how to go about it. Could you please head me in the right direction?

- Pre-Calculus -
**bobpursley**, Sunday, September 6, 2009 at 8:35pmsquare both sides.

The right side will be onerous, gather terms, combine with the right side, and you have a fourth degree polynomial that might factor.

Another way: graph y= (x-4)^2-3-sqrt(3x-4) and look for zeros. You should be able to do that on a graphical calc quickly.

- Pre-Calculus -
**MathMate**, Sunday, September 6, 2009 at 8:46pmIndeed, it is a more challenging problem.

If you keep the square root, chances are that you have to go to a numerical solution, such as Newton's method, or the bisection method to locate the roots. If you square both sides to eliminate the square root radical, and resulting in a quartic, you will need to eliminate false roots due to the squaring by back substituting the roots of the quartic.

In a case like this, it is much easier to make a sketch of both functions.

f1(x)=sqrt(3*x-4)-(x-4)^2+3

see

http://i263.photobucket.com/albums/ii157/mathmate/ashley1.png

and that of function 2 after squaring:

f2(x)=x^4-16*x^3+90*x^2+(-211)*x+173

see

http://i263.photobucket.com/albums/ii157/mathmate/ashley2.png

We can see two false roots between 3 and 5, while the roots at 1.8 and 6.6 appear in both graphs.

Now you can proceed to refine the roots at the proper locations by whichever appropriate method you choose.

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